1. Copyright © Cengage Learning. All rights reserved.12
Further Applications
of the Derivative
Copyright © Cengage Learning. All rights reserved.

2. Copyright © Cengage Learning. All rights reserved.
12.1
Maxima and Minima
Copyright © Cengage Learning. All rights reserved.

3. Maxima and Minima
Figure 1 shows the graph of a function f whose domain is the closed interval [a, b].
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.
Figure 1

4. Maxima and Minima
Figure 2 shows the graph once again with the highs and lows marked.
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.
Figure 2

5. 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.

6. 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

7. 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.

8. 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.
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.
Figure 2

9. Maxima and Minima Absolute Extrema
f has an absolute maximum at r if f (r) ≥ f (x) for every x in the domain of f.
f has an absolute minimum at 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.

10. Maxima and Minima
Some graphs have no absolute extrema at all (think of the graph of y = x), while others might have an absolute maximum but no absolute minimum (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.
(See Figure 5.)
Absolute maxima at x = a and x = b
Figure 5

11. Maxima and Minima
In Figure 6 we see the graph from Figure 1 once more, but we have labeled each extreme point as one of three types.
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.
Figure 1
Figure 6

12. 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 6 at x = q, but there is no horizontal tangent there. In fact, there is no tangent line at all; f (q) is not defined. (Recall a similar situation with the graph of f (x) = | x | at x = 0.)

13. Maxima and Minima
When f (x) does not exist for some 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.

14. 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 7:
Figure 7

15. Maxima and Minima Locating Candidates for Relative Extrema
If f is a real valued function, then its relative extrema occur among the following types of points:
Stationary Points: f has a stationary point at x if x is in 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 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.

16. Maxima and Minima
3. Endpoints: The x-coordinates of endpoints are endpoints of the domain, if any. Recall that closed intervals contain endpoints, but open intervals do not.
Once we have the x-coordinates of a candidate for a relative extremum, we find the corresponding y-coordinate using y = f (x).

17. Maxima and Minima Quick Examples
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.
Thus, the stationary points are (–2, f (–2)) = (–2, 16) and (2, f (2)) = (2, –16).
2. Singular Points: Let f (x) = 3(x – 1)1/ 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. Its coordinates are (1, f (1)) = (1, 0).

18. Maxima and Minima 3. Endpoints: Let f (x) = 1/x , with domain
( , 0)  [1, ). Then the only endpoint in the domain of f occurs when x = 1 and has coordinates
(1, 1). The natural domain of 1/x has no endpoints.

19. 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 domain.
Thus, the only candidate for a stationary relative extremum occurs when x = 1.

20. Example 1 – Solution
cont’d
Singular Points: We look for 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 ):

21. 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 8.
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 8

22. Example 1 – Solution
cont’d
Connecting these points must give us a graph something like that in Figure 9.
Notice that the graph has a horizontal tangent line at x = 1 but not at either of the endpoints because the endpoints are not stationary points.
Figure 9

23. Example 1 – Solution
cont’d
From Figure 9 we can see that f has the following extrema:

24. First Derivative Test

25. First Derivative Test
The first derivative test gives another, very systematic, way of checking whether a critical point is a relative 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:

26. First Derivative Test
At x = 0 (to the left of the critical point) we see that f (0) = –2 < 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].

27. 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 relative minimum.
Notice how the relative minimum is suggested by the arrows to the left and right.

28. First Derivative Test First Derivative Test for Relative 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 relative 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 relative minimum at x = c.
3. If f (x) has the same sign on both sides of x = c, then f has neither a relative maximum nor a relative minimum at x = c.

29. 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:

30. 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 relative maximum at x = 1. (Notice again how this is suggested by the direction of the arrows.)

31. 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 relative extrema.

32. Example 2 – Solution Singular points:
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:

33. 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.

34. Example 2 – Solution Using a Hand Plot:
cont’d
Using a Hand Plot:
If we plot these points by hand, we obtain Figure 11(a), which suggests Figure 11(b).
We can’t be sure what happens to the right of x = 1. Does the curve go up, or does it go down?
Figure 11

35. 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 12.
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 12

36. Example 2 – Solution Using the First Derivative Test:
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 12.

37. 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.

38. 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.

39. 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.