§3. Stationary points and differential calculus. Extracurricular lesson - extremum of the function
Stationary points of a function.
A necessary condition for a local extremum of a function
The first sufficient condition for a local extremum
Second and third sufficient conditions for a local extremum
The smallest and largest values of a function on a segment
Convex functions and inflection points
1. Stationary points of the function. A necessary condition for a local extremum of a function
Definition 1 
. 
Let the function be defined on 
. Dot 
called the stationary point of the function 
, If 
.
differentiated at a point
And 
Theorem 1 (necessary condition for a local extremum of a function) 
. Let the function 
determined on
and has at the point
local extremum. Then one of the conditions is satisfied:
Thus, in order to find points that are suspicious for an extremum, it is necessary to find stationary points of the function and points at which the derivative of the function does not exist, but which belong to the domain of definition of the function. 
Example 
. Let
. Find points for it that are suspicious for extremum. To solve the problem, first of all, we find the domain of definition of the function: 
. Let us now find the derivative of the function:
Points at which the derivative does not exist: 
. Stationary function points: 
Since and
, And
belong to the domain of definition of the function, then both of them will be suspicious for an extremum. But in order to conclude whether there will really be an extremum there, it is necessary to apply sufficient conditions for the extremum.
And 
Theorem 1 (necessary condition for a local extremum of a function) 
2. The first sufficient condition for a local extremum 
Theorem 1 (first sufficient condition for local extremum) 
and differentiated on this interval everywhere except, perhaps, the point 
, but at this point 
function 
is continuous. If there are such right and left semi-neighborhoods of a point
, in each of which 
retains a certain sign, then 
. Dot 
1) function
has a local extremum at the point 
takes values of different signs in the corresponding semi-neighborhoods; 
2) function 
does not have a local extremum at the point
, if to the right and left of the point
has the same sign. 
Proof 
. 1) Suppose that in a semi-neighborhood 
.
derivative 
and differentiated on this interval everywhere except, perhaps, the point 
, and in
So at the point 
has a local extremum, namely a local maximum, which was what needed to be proved. 
2) Suppose that to the left and right of the point 
, If 
and differentiated on this interval everywhere except, perhaps, the point 
the derivative retains its sign, for example,
Thus the extremum at the point 
and differentiated on this interval everywhere except, perhaps, the point 
does not have, which was what needed to be proven.
Note 1
. If the derivative 
when passing through a point 
changes sign from “+” to “-”, then at the point 
and differentiated on this interval everywhere except, perhaps, the point 
has a local maximum, and if the sign changes from “-” to “+”, then it has a local minimum.
Note 2
. An important condition is the continuity of the function 
at the point 
. If this condition is not met, then Theorem 1 may not hold.
local extremum. Then one of the conditions is satisfied: . The function is considered (Fig. 1):
This function is defined on 
and is continuous everywhere except a point 
, where it has a removable gap. When passing through a point 
changes sign from “-” to “+”, but the function does not have a local minimum at this point, but has a local maximum by definition. Indeed, near the point 
it is possible to construct a neighborhood such that for all arguments from this neighborhood the function values will be less than the value 
. Theorem 1 did not work because at the point 
the function had a gap.
Note 3
. The first sufficient condition for a local extremum cannot be used when the derivative of the function 
changes its sign in each left and each right semi-neighborhood of a point 
.
local extremum. Then one of the conditions is satisfied: . The function being considered is:
Because the 
, That 
, and therefore 
, But 
. Thus:
,
those. at the point 
and differentiated on this interval everywhere except, perhaps, the point 
has a local minimum by definition. Let's see if the first sufficient condition for a local extremum works here.
For 
:
For the first term on the right side of the resulting formula we have:
,
and therefore in a small neighborhood of the point 
the sign of the derivative is determined by the sign of the second term, that is:
,
which means that in any neighborhood of the point 
will take both positive and negative values. Indeed, consider an arbitrary neighborhood of the point 
:
. When
,
That 
(Fig. 2), and 
changes its sign here infinitely many times. Thus, the first sufficient condition for a local extremum cannot be used in the example given.
In the previous discussions we did not use the technical methods of differential calculus at all.
It is difficult not to admit that our elementary methods are simpler and more direct than the methods of analysis. In general, when dealing with a particular scientific problem, it is better to proceed from its individual characteristics than to rely solely on general methods, although, on the other hand, the general principle that clarifies the meaning of the special procedures used, of course, should always play a guiding role. This is precisely the significance of the methods of differential calculus when considering extremal problems. The desire for generality observed in modern science represents only one side of the matter, since what is truly vital in mathematics is, without any doubt, determined by the individual characteristics of the problems considered and the methods used.
In its historical development, differential calculus was to a very significant extent influenced by individual problems associated with finding the largest and smallest values of quantities. The connection between extremal problems and differential calculus can be understood as follows. In Chapter VIII we will engage in a detailed study of the derivative f"(x) of the function f(x) and its geometric meaning. There we will see that, briefly speaking, the derivative f"(x) is the slope of the tangent to the curve y = f(x) at point (x, y). It is geometrically obvious that at the maximum or minimum points of a smooth curve y = f(x) the tangent to the curve must certainly be horizontal, i.e., the slope must be zero. Thus, we obtain the condition for extremum points f"(x) = 0.
To clearly understand what it means for the derivative f"(x) to vanish, consider the curve shown in Fig. 191. We see here five points A, B, C, D, ?, at which the tangent to the curve is horizontal ; let us denote the corresponding values of f(x) at these points by a, b, c, d, e. The largest value of f(x) (within the area shown in the drawing) is achieved at point D, the smallest at point A. At point B there is a maximum - in the sense that at all points some neighborhood points B, the value of f(x) is less than b, although at points close to D, the value of f(x) is still greater than b. For this reason, it is customary to say that at point B there is relative maximum of function f(x), whereas at point D - absolute maximum. In the same way, at point C there is relative minimum, and at point A - absolute minimum. Finally, as for point E, there is neither a maximum nor a minimum in it, although the equality is still realized in it f"(x) = Q, It follows that the vanishing of the derivative f"(x) is necessary, but not at all sufficient condition for the appearance of an extremum of a smooth function f(x); in other words, at any point where there is an extremum (absolute or relative), the equality certainly takes place f"(x) = 0, but not at every point where f"(x) = 0, must be an extremum. Those points at which the derivative f"(x) vanishes, regardless of whether there is an extremum at them, are called stationary. Further analysis leads to more or less complex conditions concerning the higher derivatives of the function f(x) and completely characterizing the maxima, minima and other stationary points.
Definitions:
Extremum call the maximum or minimum value of a function on a given set.
Extremum point is the point at which the maximum or minimum value of the function is reached.
Maximum point is the point at which the maximum value of the function is reached.
Minimum point is the point at which the minimum value of the function is reached.
Explanation.
In the figure, in the vicinity of the point x = 3, the function reaches its maximum value (that is, in the vicinity of this particular point there is no point higher). In the neighborhood of x = 8, it again has a maximum value (let us clarify again: it is in this neighborhood that there is no point higher). At these points, the increase gives way to a decrease. They are the maximum points:
x max = 3, x max = 8.
In the vicinity of the point x = 5, the minimum value of the function is reached (that is, in the vicinity of x = 5 there is no point below). At this point the decrease gives way to an increase. It is the minimum point:
The maximum and minimum points are extremum points of the function, and the values of the function at these points are its extremes.
Critical and stationary points of the function:
Necessary condition for an extremum:
Sufficient condition for an extremum:
On a segment the function y = f(x) can reach its minimum or maximum value either at critical points or at the ends of the segment .
Algorithm for studying a continuous functiony = f(x) for monotonicity and extrema:
Critical points– these are the points at which the derivative of a function is equal to zero or does not exist. If the derivative is equal to 0 then the function at this point takes local minimum or maximum. On the graph at such points the function has a horizontal asymptote, that is, the tangent is parallel to the Ox axis.
Such points are called stationary. If you see a “hump” or “hole” on the graph of a continuous function, remember that the maximum or minimum is reached at a critical point. Let's take the following task as an example.
Example 1. Find the critical points of the function y=2x^3-3x^2+5.
Solution. The algorithm for finding critical points is as follows:
So the function has two critical points.
Next, if you need to study a function, then we determine the sign of the derivative to the left and to the right of the critical point. If the derivative changes sign from “-” to “+” when passing through the critical point, then the function takes local minimum. If from “+” to “-” should local maximum.
Second type of critical points these are the zeros of the denominator of fractional and irrational functions
Logarithmic and trigonometric functions that are not defined at these points 
Third type of critical points have piecewise continuous functions and modules.
For example, any module-function has a minimum or maximum at the break point.
For example module y = | x -5 |
at point x = 5 has a minimum (critical point).
The derivative does not exist in it, but on the right and left takes the value 1 and -1, respectively.
1)
2)
3)
4)
5)
Try to determine the critical points of functions
If the answer is y you get the value
1) x=4;
2) x=-1;x=1;
3) x=9;
4) x=Pi*k;
5) x=1. then you already know how to find critical points
and be able to cope with a simple test or tests.
The process of examining a function for the presence of stationary points and also finding them is one of the important elements when constructing a graph of a function. You can find stationary points of a function if you have a certain set of mathematical knowledge.
- You will need
- - function that needs to be examined for the presence of stationary points;
- definition of stationary points: stationary points of a function are points (argument values) at which the derivative of the first order function vanishes.
- Instructions
- Using the table of derivatives and formulas for differentiating functions, it is necessary to find the derivative of the function. This step is the most difficult and critical in completing the task. If you make a mistake at this stage, further calculations will not make sense.
- Compose the equation f"(x) = 0 and solve it. The equation may have no solutions - in this case, the function does not have stationary points. If the equation has solutions, then these particular values of the argument will be the stationary points of the function. At this point At this stage, the solution to the equation should be checked by argument substitution.
