As a matter of terminology, from now on we shall refer to the derivative of a function alternatively as its first derivative (short fox first-order derivative). The reason for this will become apparent shortly.

Given a function y = f(x)> the first derivative /'(v) plays a major role in our search for its extreme values. This is due to the fact that, if a relative extremum of the function occurs at x — Xi)t then either (1) /'{.to) does not exist, or (2) /'(-To) = 0. The first eventuality is illustrated in Fig. 9.2a, where both points A and B depict relative extreme values of v, and yet no derivative is defined at either of these sharp points. Since in the present discussion we are assuming that y = f(x) is continuous and possesses a continuous derivative, however, we are in effect ruling out sharp points. For smooth functions, relative extreme values can occur only where the first derivative has a zero value. This is illustrated by points C and D in Fig, 9.2b, both of which represent extreme values, and both of which are characterized by a zero slope—ff(x\) — 0 and f\x^) = 0. It is also easy to see that when the slope is nonzero we cannot possibly have a relative minimum (the bottom of a valley) or a relative maximum (the peak of a hill). For this reason, we can, in the context of smooth functions, take the condition f{x) = 0 to be a necessary condition for a relative extremum (either maximum or minimum).

We must hasten to add, however, that a zero slope, while necessary; is not sufficient to establish a relative extremum. An example of the ease where a zero slope is not associated with an extremum will be presented shortly. By appending a certain proviso to the zero-slope condition, however, we can obtain a decisive test for a relative extremum. This may be stated as follows:

First-derivative test for relative extremum If the first derivative of a function/(r) at x — Jto is f(xo) — 0, then the value of the function at*0; /{-to), will be a. A relative maximum if the derivative f'(x) changes its sign from positive to negative from the immediate left of the point to its immediate right.

b, A relative minimum if f{x) changes its sign from negative to positive from the immediate left of xo to its immediate right.

c. Neither a relative maximum nor a relative minimum if f\x) has the same sign on both the immediate left and the immediate right of point xq.

Let us call the value xq a critical value of x if f(xo) = 0, and refer to f(xo) as a stationary value of y (or of the function/). The point with coordinates and /(*o) can, accordingly, be called a stationary point (The rationale for the word stationary should be self-evident—wherever the slope is zero, the point in question is never situated on an upward or downward incline, but is rather at a standstill position.) Then, graphically, the first possibility listed in this test will establish the stationary point as the peak of a hill, such as point D in Fig. 92b, whereas the second possibility will establish the stationary point as the bottom of a valley, such as point C in the same diagram. Note, however, that in view of the existence of a third possibility, yet to be discussed, we are unable to regard the condition f(x) = 0 as a sufficient condition for a relative extremum. But we now see that, /fthe necessary condition f'{x) = 0 is satisfied, then the change-of-derivative-sign proviso can serve as a sufficient condition for a relative maximum or minimum, depending on the direction of tne sign change.

Let us now explain the third possibility. In Fig, 9.3a, the function/is shown to attain a zero slope at point J (when x = j\ Even though f(j) is zero—which makes f(j) a stationary value—the derivative does not change its sign from one side of x = j to the other; therefore, according to the first-derivative test, point J gives neither a maximum nor

FIGURE 9.3

FIGURE 9.3

dl dx

x a minimum, as is duly confirmed by the graph of the function. Rather, it exemplifies what is known as an inflection point.

The characteristic feature of an inflection point is that, at that point, the derivative (as against the primitive) function reaches an extreme value. Since this extreme value can be either a maximum or a minimum, we have two types of inflection points. In Fig. 9.3a', where we have plotted the derivative /'(r), we see that its value is zero when x = j (see point J') but is positive on both sides of point .V\ this makes f a minimum point of the derivative function f(x)t

The other type of inflection point is portrayed in Fig, 93i>> where the slope of the function g{x) increases till the point k is reached and decreases thereafter. Consequently, the graph of the derivative function g'(x) will assume the shape shown in Fig. 9.3b', where point K! gives a maximum value of the derivative function g'(x)*

To sum up: A relative exlremum must be a stationary value, but a stationary value may be associated with either a relative extremum or an inflection point. To find the relative maximum or minimum of a given function, therefore, the procedure should be first to find the stationary values of the function where the condition ff(x) — 0 is satisfied, and then to apply the first-derivative test to determine whether each of the stationary values is a relative maximum, a relative minimum, or neither.

Find the relative extrema of the function y= = x3 - -h 36x + 8 First, we find the derivative function to be f'(x) = lx2 - 24x + 36

To get the critical values, i.e., the values of x satisfying the condition f'(x) = 0, we set the quadratic derivative function equal to zero and get the quadratic equation

By factoring the polynomial or by applying the quadratic formula, we then obtain the following pair of roots (solutions):

= 6 [at which we have f(6) = 0 and f(6) = 8] x* = 2 [at which we have f\2) = 0 and f(2) = 40]

Since f*{6) = f (2) = 0, these two values of x are the critical values we desire,

It is easy to verify that, in the immediate neighborhood of* =6, we have f{x) < 0 for x < 6, and ff(x) > 0 for x >6; thus the value of the function /(6) = 8 is a relative minimum. Similarly, since, in the immediate neighborhood of x = 2, we find f(x) > 0 for x <2, and /'(*) <0 for x > 2, the value of the function /(2) = 40 is a relative maximum.

T Note that a zero derivative value, while a necessary condition for a retative extremum, is not required for an inflection point; for the derivative g'(x) has a positive value at x = k, and yet point K is an inflection point.

Example 1

FIGURE 9,4

Figure 9.4 shows the graph of the function of this example. Such a graph may be used to verify the location of extreme values obtained through use of the first-derivative test. But, in reality, in most cases "helpfulness" flows in the opposite direction—the mathematically derived extreme values will help in plotting the graph, The accurate plotting of a graph ideally requires knowledge of the value of the function at every point in the domain; but as a matter of actual practice, only a few points in the domain are selected for purposes of plotting, and the rest of the points typically are filled in by interpolation. The pitfall of this practice is that, unless we hit upon the stationary point(s) by coincidence, we shall miss the exact location of the turning point(s) in the curve. Now, with the first-derivative test at our disposal, it becomes possible to locate these turning points precisely.

Figure 9.4 shows the graph of the function of this example. Such a graph may be used to verify the location of extreme values obtained through use of the first-derivative test. But, in reality, in most cases "helpfulness" flows in the opposite direction—the mathematically derived extreme values will help in plotting the graph, The accurate plotting of a graph ideally requires knowledge of the value of the function at every point in the domain; but as a matter of actual practice, only a few points in the domain are selected for purposes of plotting, and the rest of the points typically are filled in by interpolation. The pitfall of this practice is that, unless we hit upon the stationary point(s) by coincidence, we shall miss the exact location of the turning point(s) in the curve. Now, with the first-derivative test at our disposal, it becomes possible to locate these turning points precisely.

Example 2

Find the relative extremum of the average-cost function

The derivative here is T(Q) = 2Q - 5, a linear function. Setting f'(Q) equal to zero, we get the linear equation 2Q - 5 = 0, which has the single root Q* = 2.5, This is the only critical value in this case, To apply the first-derivative test, let us find the values of the derivative at, say, Q = 2.4 and Q = 2.6, respectively. Since /'(2.4) = -0.2 <0 whereas f'{2.6) = 0.2 > 0, we can conclude that the stationary value AC = f(2.5) = 1,75 represents a relative minimum. The graph of the function of this example is actually a U-shaped curve, so that the relative minimum already found will also be the absolute minimum, Our knowledge of the exact location of this point should be of great help in plotting the AC curve.

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