An important concept in mathematics is a function. With its help, you can visualize many processes occurring in nature, reflect the relationship between certain quantities using formulas, tables and images on a graph. An example is the dependence of the pressure of a liquid layer on a body on the depth of immersion, acceleration - on the action of a certain force on an object, temperature increase - on the transmitted energy, and many other processes. The study of a function involves plotting a graph, finding out its properties, the domain of definition and values, intervals of increase and decrease. An important point in this process is finding the extremum points. About how to do it right, and the conversation will go on.
About the concept itself on a specific example
In medicine, the construction of a function graph can tell about the course of the development of the disease in the patient's body, clearly reflecting his condition. Let's assume that the time in days is plotted along the OX axis, and the temperature of the human body is plotted along the OY axis. The figure clearly shows how this indicator rises sharply, and then falls. It is also easy to notice singular points that reflect the moments when the function, having previously increased, begins to decrease, and vice versa. These are the extreme points, that is, the critical values (maximum and minimum) in this case of the patient's temperature, after which changes in his condition occur.
Tilt angle
It is easy to determine from the figure how the derivative of the function changes. If the straight lines of the graph go up over time, then it is positive. And the steeper they are, the greater the value of the derivative, as the angle of inclination increases. During periods of decrease, this value takes on negative values, turning to zero at extremum points, and the graph of the derivative in the latter case is drawn parallel to the OX axis.
Any other process should be treated in the same way. But the best way to tell about this concept is the movement of various bodies, clearly shown on the graphs.
Movement
Suppose some object moves in a straight line, gaining speed uniformly. During this period, the change in the coordinates of the body graphically represents a certain curve, which a mathematician would call a branch of a parabola. At the same time, the function is constantly increasing, since the coordinate indicators change faster and faster with every second. The speed graph shows the behavior of the derivative, the value of which also increases. This means that the movement has no critical points.
This would continue indefinitely. But what if the body suddenly decides to slow down, stop and start moving in a different direction? In this case, the coordinate indicators will begin to decrease. And the function will pass a critical value and turn from increasing into decreasing.
In this example, you can again understand that the extremum points on the graph of the function appear at the moments when it ceases to be monotonous.
The physical meaning of the derivative
What was described earlier clearly showed that the derivative is essentially the rate of change of the function. This refinement contains its physical meaning. Extreme points are critical areas on the chart. It is possible to find out and detect them by calculating the value of the derivative, which turns out to be equal to zero.
There is another sign, which is a sufficient condition for an extremum. The derivative in such places of inflection changes its sign: from "+" to "-" in the region of the maximum and from "-" to "+" in the region of the minimum.
Movement under the influence of gravity
Let's imagine another situation. The children, playing ball, threw it in such a way that it began to move at an angle to the horizon. At the initial moment, the speed of this object was the largest, but under the influence of gravity, it began to decrease, and with each second by the same value, equal to approximately 9.8 m / s 2. This is the value of the acceleration that occurs under the influence of the earth's gravity during free fall. On the Moon, it would be about six times smaller.
The graph describing the movement of the body is a parabola with branches pointing downward. How to find extremum points? In this case, this is the vertex of the function, where the speed of the body (ball) takes on a zero value. The derivative of the function becomes zero. In this case, the direction, and hence the value of the speed, changes to the opposite. The body flies down with every second faster and faster, and accelerates by the same amount - 9.8 m/s 2 .
Second derivative
In the previous case, the plot of the velocity modulus is drawn as a straight line. This line is first directed downwards, since the value of this quantity is constantly decreasing. Having reached zero at one of the time points, then the indicators of this value begin to increase, and the direction of the graphical representation of the speed module changes dramatically. Now the line is pointing up.
Velocity, being the derivative of the coordinate with respect to time, also has a critical point. In this region, the function, initially decreasing, begins to increase. This is the place of the extremum point of the derivative of the function. In this case, the slope of the tangent becomes zero. And acceleration, being the second derivative of the coordinate with respect to time, changes sign from “-” to “+”. And the movement from uniformly slow becomes uniformly accelerated.
Acceleration Graph
Now consider four figures. Each of them displays a graph of the change over time of such a physical quantity as acceleration. In the case of "A", its value remains positive and constant. This means that the speed of the body, like its coordinate, is constantly increasing. If we imagine that the object will move in this way for an infinitely long time, the function reflecting the dependence of the coordinate on time will turn out to be constantly increasing. It follows from this that it has no critical regions. There are also no extremum points on the graph of the derivative, that is, a linearly changing speed.
The same applies to case "B" with a positive and constantly increasing acceleration. True, the graphs for coordinates and speed will be somewhat more complicated here.
When acceleration goes to zero
Looking at figure "B", one can observe a completely different picture characterizing the movement of the body. Its speed will be graphically depicted as a parabola with branches pointing downwards. If we continue the line describing the change in acceleration until it intersects with the OX axis, and further, then we can imagine that up to this critical value, where the acceleration turns out to be equal to zero, the speed of the object will increase more and more slowly. The extremum point of the derivative of the coordinate function will be just at the top of the parabola, after which the body will radically change the nature of the movement and begin to move in a different direction.
In the latter case, "G", the nature of the movement cannot be precisely determined. Here we only know that there is no acceleration for some period under consideration. This means that the object can remain in place or the movement occurs at a constant speed.
Coordinate addition task
Let's move on to tasks that are often encountered when studying algebra at school and are offered to prepare for the exam. The figure below shows the graph of the function. It is required to calculate the sum of extremum points.
We will do this for the y-axis by determining the coordinates of the critical regions where a change in the characteristics of the function is observed. Simply put, we find the values along the x-axis for the inflection points, and then proceed to add the resulting terms. According to the graph, it is obvious that they take the following values: -8; -7; -5; -3; -2; 1; 3. This adds up to -21, which is the answer.
Optimal solution
It is not necessary to explain how important the choice of the optimal solution can be in the performance of practical tasks. After all, there are many ways to achieve the goal, and the best way out, as a rule, is only one. This is extremely necessary, for example, when designing ships, spacecraft and aircraft, architectural structures in order to find the optimal form of these man-made objects.
The speed of vehicles largely depends on the competent minimization of the resistance that they experience when moving through water and air, on overloads arising under the influence of gravitational forces and many other indicators. A ship at sea needs such qualities as stability during a storm; for a river ship, a minimum draft is important. When calculating the optimal design, the extremum points on the graph can visually give an idea of the best solution to a complex problem. The tasks of such a plan are often solved in the economy, in economic areas, in many other life situations.
From ancient history
Extreme tasks occupied even the ancient sages. Greek scientists successfully unraveled the mystery of areas and volumes through mathematical calculations. They were the first to understand that on a plane of various figures with the same perimeter, the circle always has the largest area. Similarly, a ball is endowed with the maximum volume among other objects in space with the same surface area. Such famous personalities as Archimedes, Euclid, Aristotle, Apollonius devoted themselves to solving such problems. Heron succeeded very well in finding extremum points, who, having resorted to calculations, built ingenious devices. These included automatic machines moving by means of steam, pumps and turbines operating on the same principle.
Construction of Carthage
There is a legend, the plot of which is based on solving one of the extreme tasks. The result of the business approach demonstrated by the Phoenician princess, who turned to the sages for help, was the construction of Carthage. The land plot for this ancient and famous city was presented to Dido (that was the name of the ruler) by the leader of one of the African tribes. The area of the allotment did not seem to him at first very large, since according to the contract it had to be covered with an oxhide. But the princess ordered her soldiers to cut it into thin strips and make a belt out of them. It turned out to be so long that it covered an area where the whole city fit.
Origins of calculus
And now let's move from ancient times to a later era. Interestingly, in the 17th century, Kepler was prompted to understand the foundations of mathematical analysis by a meeting with a wine seller. The merchant was so well versed in his profession that he could easily determine the volume of the drink in the barrel by simply lowering an iron tourniquet into it. Reflecting on such a curiosity, the famous scientist managed to solve this dilemma for himself. It turns out that skillful coopers of those times got the hang of making vessels in such a way that, at a certain height and radius of the circumference of the fastening rings, they would have a maximum capacity.
This became for Kepler an occasion for further reflection. Bochars came to the optimal solution by a long search, mistakes and new attempts, passing their experience from generation to generation. But Kepler wanted to speed up the process and learn how to do the same in a short time through mathematical calculations. All his developments, picked up by colleagues, turned into the now known theorems of Fermat and Newton - Leibniz.
The problem of finding the maximum area
Imagine that we have a wire whose length is 50 cm. How to make a rectangle out of it, which has the largest area?
Starting a decision, one should proceed from simple and well-known truths. It is clear that the perimeter of our figure will be 50 cm. It also consists of twice the lengths of both sides. This means that, having designated one of them as "X", the other can be expressed as (25 - X).
From here we get an area equal to X (25 - X). This expression can be represented as a function that takes on many values. The solution of the problem requires finding the maximum of them, which means that you should find out the extremum points.
To do this, we find the first derivative and equate it to zero. The result is a simple equation: 25 - 2X = 0.
From it we learn that one of the sides is X = 12.5.
Therefore, another: 25 - 12.5 \u003d 12.5.
It turns out that the solution to the problem will be a square with a side of 12.5 cm.
How to find the maximum speed
Let's consider one more example. Imagine that there is a body whose rectilinear motion is described by the equation S = - t 3 + 9t 2 - 24t - 8, where the distance traveled is expressed in meters, and the time in seconds. It is required to find the maximum speed. How to do it? Downloaded find the speed, that is, the first derivative.
We get the equation: V = - 3t 2 + 18t - 24. Now, to solve the problem, we again need to find the extremum points. This must be done in the same way as in the previous task. We find the first derivative of the speed and equate it to zero.
We get: - 6t + 18 = 0. Hence t = 3 s. This is the time when the speed of the body takes on a critical value. We substitute the obtained data into the velocity equation and get: V = 3 m/s.
But how to understand that this is exactly the maximum speed, because the critical points of the function can be its largest or smallest values? To check, you need to find the second derivative of the speed. It is expressed as the number 6 with a minus sign. This means that the found point is the maximum. And in the case of a positive value of the second derivative, there would be a minimum. Hence, the solution found was correct.
The tasks given as an example are only a part of those that can be solved by being able to find the extremum points of a function. In fact, there are many more. And such knowledge opens up unlimited possibilities for human civilization.
Consider two teeth of a well-known saw profile. Let's direct the axis along the flat side of the saw, and the axis - perpendicular to it. Let's get a graph of some function, shown in Fig. 1.
It is quite obvious that both at the point and at the point, the values of the function turn out to be the largest in comparison with the values at the neighboring points on the right and left, and at the point - the smallest in comparison with the neighboring points. The points are called the extremum points of the function (from the Latin extremum - “extreme”), the points and are the maximum points, and the point is the minimum point (from the Latin maximum and minimum - “greatest” and “smallest”).
Let us refine the definition of an extremum.
A function at a point is said to have a maximum if there is an interval containing the point and belonging to the domain of the function, such that for all points of this interval it turns out to be . Accordingly, the function at a point has a minimum if the condition is satisfied for all points of a certain interval.
On fig. Figures 2 and 3 show graphs of functions that have an extremum at a point.
Let us pay attention to the fact that, by definition, the extremum point must lie inside the interval of setting the function, and not at its end. Therefore, for the function shown in Fig. 1, it cannot be assumed that it has a minimum at the point.
If in this definition of the maximum (minimum) of a function, we replace the strict inequality with a non-strict one 
, then we obtain the definition of a non-strict maximum (non-strict minimum). Consider, for example, the profile of a mountain top (Fig. 4). Each point of a flat area - a segment is a non-strict maximum point.
In differential calculus, the study of a function for extrema is very effective and quite simply carried out using a derivative. One of the main theorems of differential calculus, which establishes a necessary condition for the extremum of a differentiable function, is Fermat's theorem (see Fermat's theorem). Let the function at a point have an extremum. If there is a derivative at this point, then it is equal to zero.
In geometric language, Fermat's theorem means that at the extremum point the tangent to the graph of the function is horizontal (Fig. 5). The converse statement, of course, is not true, which is shown, for example, by the graph in Fig. 6.
The theorem is named after the French mathematician P. Fermat, who was one of the first to solve a number of extremum problems. He did not yet have the concept of a derivative at his disposal, but in his investigation he used a method whose essence is expressed in the statement of the theorem.
A sufficient condition for the extremum of a differentiable function is a change in the sign of the derivative. If at a point the derivative changes sign from minus to plus, i.e. its decrease is replaced by an increase, then the point will be the minimum point. On the contrary, the point will be the maximum point if the derivative changes sign from plus to minus, i.e. goes from ascending to descending.
The point where the derivative of the function is equal to zero is called stationary. If a differentiable function is being investigated for an extremum, then all its stationary points should be found and the signs of the derivative should be considered to the left and to the right of them.
We investigate the function for an extremum.
Let's find its derivative: 
.
Let us turn to the graph of the function y \u003d x 3 - 3x 2. Consider the neighborhood of the point x = 0, i.e. some interval containing this point. It is logical that there is such a neighborhood of the point x \u003d 0 that the function y \u003d x 3 - 3x 2 takes the largest value in this neighborhood at the point x \u003d 0. For example, on the interval (-1; 1) the largest value equal to 0, the function takes at the point x = 0. The point x = 0 is called the maximum point of this function.
Similarly, the point x \u003d 2 is called the minimum point of the function x 3 - 3x 2, since at this point the value of the function is not greater than its value at another point in the vicinity of the point x \u003d 2, for example, the neighborhood (1.5; 2.5).
Thus, the point x 0 is called the maximum point of the function f (x) if there is a neighborhood of the point x 0 - such that the inequality f (x) ≤ f (x 0) is satisfied for all x from this neighborhood.
For example, the point x 0 \u003d 0 is the maximum point of the function f (x) \u003d 1 - x 2, since f (0) \u003d 1 and the inequality f (x) ≤ 1 is true for all values of x.
The minimum point of the function f (x) is called the point x 0 if there is such a neighborhood of the point x 0 that the inequality f (x) ≥ f (x 0) is satisfied for all x from this neighborhood.
For example, the point x 0 \u003d 2 is the minimum point of the function f (x) \u003d 3 + (x - 2) 2, since f (2) \u003d 3 and f (x) ≥ 3 for all x.
Extreme points are called minimum points and maximum points.
Let us turn to the function f(x), which is defined in some neighborhood of the point x 0 and has a derivative at this point.
If x 0 is an extremum point of a differentiable function f (x), then f "(x 0) \u003d 0. This statement is called Fermat's theorem.
Fermat's theorem has a clear geometric meaning: at the extremum point, the tangent is parallel to the x-axis and therefore its slope
f "(x 0) is zero.
For example, the function f (x) \u003d 1 - 3x 2 has a maximum at the point x 0 \u003d 0, its derivative f "(x) \u003d -2x, f "(0) \u003d 0.
The function f (x) \u003d (x - 2) 2 + 3 has a minimum at the point x 0 \u003d 2, f "(x) \u003d 2 (x - 2), f "(2) \u003d 0.
Note that if f "(x 0) \u003d 0, then this is not enough to assert that x 0 is necessarily the extremum point of the function f (x).
For example, if f (x) \u003d x 3, then f "(0) \u003d 0. However, the point x \u003d 0 is not an extremum point, since the function x 3 increases on the entire real axis.
So, the extremum points of a differentiable function must be sought only among the roots of the equation
f "(x) \u003d 0, but the root of this equation is not always an extremum point.
Stationary points are points at which the derivative of a function is equal to zero.
Thus, in order for the point x 0 to be an extremum point, it is necessary that it be a stationary point.
Consider sufficient conditions for a stationary point to be an extremum point, i.e. conditions under which a stationary point is a minimum or maximum point of a function.
If the derivative to the left of the stationary point is positive, and to the right it is negative, i.e. derivative changes sign "+" to sign "-" when passing through this point, then this stationary point is the maximum point.
Indeed, in this case, to the left of the stationary point, the function increases, and to the right, it decreases, i.e. this point is the maximum point.
If the derivative changes sign "-" to sign "+" when passing through a stationary point, then this stationary point is a minimum point.
If the derivative does not change sign when passing through a stationary point, i.e. the derivative is positive or negative to the left and to the right of the stationary point, then this point is not an extremum point.
Let's consider one of the problems. Find the extremum points of the function f (x) \u003d x 4 - 4x 3.
Solution.
1) Find the derivative: f "(x) \u003d 4x 3 - 12x 2 \u003d 4x 2 (x - 3).
2) Find stationary points: 4x 2 (x - 3) \u003d 0, x 1 \u003d 0, x 2 \u003d 3.
3) Using the interval method, we establish that the derivative f "(x) \u003d 4x 2 (x - 3) is positive for x\u003e 3, negative for x< 0 и при 0 < х < 3.
4) Since when passing through the point x 1 \u003d 0, the sign of the derivative does not change, this point is not an extremum point.
5) The derivative changes the sign "-" to the sign "+" when passing through the point x 2 \u003d 3. Therefore, x 2 \u003d 3 is the minimum point.
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Definitions:
extremum name the maximum or minimum value of a function on a given set.
extreme 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.
Low 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 higher point). In the neighborhood of x = 8, it again has a maximum value (again, let's clarify: it is in this neighborhood that there is no point above). At these points, the increase is replaced by a decrease. They are maximum points:
xmax = 3, xmax = 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 is replaced by 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 the 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:
Let the function $z=f(x,y)$ be defined in some neighborhood of the point $(x_0,y_0)$. It is said that $(x_0,y_0)$ is a point of (local) maximum if for all points $(x,y)$ in some neighborhood of $(x_0,y_0)$ the inequality $f(x,y)< f(x_0,y_0)$. Если же для всех точек этой окрестности выполнено условие $f(x,y)>f(x_0,y_0)$, then the point $(x_0,y_0)$ is called a (local) minimum point.
High and low points are often referred to by the generic term extremum points.
If $(x_0,y_0)$ is a maximum point, then the value of the function $f(x_0,y_0)$ at this point is called the maximum of the function $z=f(x,y)$. Accordingly, the value of the function at the minimum point is called the minimum of the function $z=f(x,y)$. The minima and maxima of a function are united by a common term - the extrema of a function.
Algorithm for studying the function $z=f(x,y)$ for an extremum
- Find the partial derivatives of $\frac(\partial z)(\partial x)$ and $\frac(\partial z)(\partial y)$. Compose and solve the system of equations $ \left \( \begin(aligned) & \frac(\partial z)(\partial x)=0;\\ & \frac(\partial z)(\partial y)=0. \ end(aligned) \right.$ Points whose coordinates satisfy the specified system are called stationary.
- Find $\frac(\partial^2z)(\partial x^2)$, $\frac(\partial^2z)(\partial x\partial y)$, $\frac(\partial^2z)(\partial y^2)$ and compute the value $\Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac (\partial^2z)(\partial x\partial y) \right)^2$ at every stationary point. After that, use the following scheme:
- If $\Delta > 0$ and $\frac(\partial^2z)(\partial x^2) > 0$ (or $\frac(\partial^2z)(\partial y^2) > 0$), then at the point under study is the minimum point.
- If $\Delta > 0$ and $\frac(\partial^2z)(\partial x^2)< 0$ (или $\frac{\partial^2z}{\partial y^2} < 0$), то в исследуемая точка есть точкой максимума.
- If $\Delta< 0$, то в расматриваемой стационарной точке экстремума нет.
- If $\Delta = 0$, then nothing definite can be said about the presence of an extremum; additional research is required.
Note (desirable for a better understanding of the text): show\hide
If $\Delta > 0$ then $\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\ partial^2z)(\partial x\partial y) \right)^2 > 0$. And from this it follows that $\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2) > \left(\frac(\partial^2z) (\partial x\partial y) \right)^2 ≥ 0$. Those. $\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2) > 0$. If the product of some quantities is greater than zero, then these quantities have the same sign. That is, for example, if $\frac(\partial^2z)(\partial x^2) > 0$, then $\frac(\partial^2z)(\partial y^2) > 0$. In short, if $\Delta > 0$ then the signs of $\frac(\partial^2z)(\partial x^2)$ and $\frac(\partial^2z)(\partial y^2)$ are the same.
Example #1
Investigate the function $z=4x^2-6xy-34x+5y^2+42y+7$ for an extremum.
$$ \frac(\partial z)(\partial x)=8x-6y-34; \frac(\partial z)(\partial y)=-6x+10y+42. $$
$$ \left \( \begin(aligned) & 8x-6y-34=0;\\ & -6x+10y+42=0. \end(aligned) \right. $$
Let's reduce each equation of this system by $2$ and transfer the numbers to the right-hand sides of the equations:
$$ \left \( \begin(aligned) & 4x-3y=17;\\ & -3x+5y=-21. \end(aligned) \right. $$
We have obtained a system of linear algebraic equations. In this situation, it seems to me the most convenient application of Cramer's method to solve the resulting system.
$$ \begin(aligned) & \Delta=\left| \begin(array) (cc) 4 & -3\\ -3 & 5 \end(array)\right|=4\cdot 5-(-3)\cdot (-3)=20-9=11;\ \ & \Delta_x=\left| \begin(array) (cc) 17 & -3\\ -21 & 5 \end(array)\right|=17\cdot 5-(-3)\cdot (-21)=85-63=22;\ \ & \Delta_y=\left| \begin(array) (cc) 4 & 17\\ -3 & -21 \end(array)\right|=4\cdot (-21)-17\cdot (-3)=-84+51=-33 .\end(aligned) \\ x=\frac(\Delta_(x))(\Delta)=\frac(22)(11)=2; \; y=\frac(\Delta_(y))(\Delta)=\frac(-33)(11)=-3. $$
The values $x=2$, $y=-3$ are the coordinates of the stationary point $(2;-3)$.
$$ \frac(\partial^2 z)(\partial x^2)=8; \frac(\partial^2 z)(\partial y^2)=10; \frac(\partial^2 z)(\partial x \partial y)=-6. $$
Let's calculate the value of $\Delta$:
$$ \Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\partial^2z)( \partial x\partial y) \right)^2= 8\cdot 10-(-6)^2=80-36=44. $$
Since $\Delta > 0$ and $\frac(\partial^2 z)(\partial x^2) > 0$, then according to the point $(2;-3)$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $(2;-3)$ into the given function:
$$ z_(min)=z(2;-3)=4\cdot 2^2-6\cdot 2 \cdot (-3)-34\cdot 2+5\cdot (-3)^2+42\ cdot(-3)+7=-90. $$
Answer: $(2;-3)$ - minimum point; $z_(min)=-90$.
Example #2
Investigate the function $z=x^3+3xy^2-15x-12y+1$ for an extremum.
We will follow the above. First, let's find the partial derivatives of the first order:
$$ \frac(\partial z)(\partial x)=3x^2+3y^2-15; \frac(\partial z)(\partial y)=6xy-12. $$
Compose the system of equations $ \left \( \begin(aligned) & \frac(\partial z)(\partial x)=0;\\ & \frac(\partial z)(\partial y)=0. \end( aligned)\right.$:
$$ \left \( \begin(aligned) & 3x^2+3y^2-15=0;\\ & 6xy-12=0. \end(aligned) \right. $$
Reduce the first equation by 3 and the second by 6.
$$ \left \( \begin(aligned) & x^2+y^2-5=0;\\ & xy-2=0. \end(aligned) \right. $$
If $x=0$, then the second equation will lead us to a contradiction: $0\cdot y-2=0$, $-2=0$. Hence the conclusion: $x\neq 0$. Then from the second equation we have: $xy=2$, $y=\frac(2)(x)$. Substituting $y=\frac(2)(x)$ into the first equation, we have:
$$ x^2+\left(\frac(2)(x) \right)^2-5=0;\\ x^2+\frac(4)(x^2)-5=0;\\ x^4-5x^2+4=0. $$
We got a biquadratic equation. We make the substitution $t=x^2$ (we keep in mind that $t > 0$):
$$ t^2-5t+4=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 1 \cdot 4=9;\\ & t_1=\frac(-(- 5)-\sqrt(9))(2)=\frac(5-3)(2)=1;\\ & t_2=\frac(-(-5)+\sqrt(9))(2)= \frac(5+3)(2)=4.\end(aligned) $$
If $t=1$, then $x^2=1$. Hence we have two values of $x$: $x_1=1$, $x_2=-1$. If $t=4$, then $x^2=4$, i.e. $x_3=2$, $x_4=-2$. Remembering that $y=\frac(2)(x)$, we get:
\begin(aligned) & y_1=\frac(2)(x_1)=\frac(2)(1)=2;\\ & y_2=\frac(2)(x_2)=\frac(2)(-1 )=-2;\\ & y_3=\frac(2)(x_3)=\frac(2)(2)=1;\\ & y_4=\frac(2)(x_4)=\frac(2)( -2)=-1. \end(aligned)
So, we have four stationary points: $M_1(1;2)$, $M_2(-1;-2)$, $M_3(2;1)$, $M_4(-2;-1)$. This completes the first step of the algorithm.
Now let's get down to the algorithm. Let's find partial derivatives of the second order:
$$ \frac(\partial^2 z)(\partial x^2)=6x; \frac(\partial^2 z)(\partial y^2)=6x; \frac(\partial^2 z)(\partial x \partial y)=6y. $$
Find $\Delta$:
$$ \Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\partial^2z)( \partial x\partial y) \right)^2= 6x\cdot 6x-(6y)^2=36x^2-36y^2=36(x^2-y^2). $$
Now we will calculate the value of $\Delta$ at each of the previously found stationary points. Let's start from the point $M_1(1;2)$. At this point we have: $\Delta(M_1)=36(1^2-2^2)=-108$. Since $\Delta(M_1)< 0$, то согласно в точке $M_1$ экстремума нет.
Let's explore the point $M_2(-1;-2)$. At this point we have: $\Delta(M_2)=36((-1)^2-(-2)^2)=-108$. Since $\Delta(M_2)< 0$, то согласно в точке $M_2$ экстремума нет.
Let's examine the point $M_3(2;1)$. At this point we get:
$$ \Delta(M_3)=36(2^2-1^2)=108;\;\; \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3)=6\cdot 2=12. $$
Since $\Delta(M_3) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3) > 0$, then according to $M_3(2; 1)$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_3$ into the given function:
$$ z_(min)=z(2;1)=2^3+3\cdot 2\cdot 1^2-15\cdot 2-12\cdot 1+1=-27. $$
It remains to explore the point $M_4(-2;-1)$. At this point we get:
$$ \Delta(M_4)=36((-2)^2-(-1)^2)=108;\;\; \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_4)=6\cdot (-2)=-12. $$
Since $\Delta(M_4) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_4)< 0$, то согласно $M_4(-2;-1)$ есть точкой максимума функции $z$. Максимум функции $z$ найдём, подставив в заданную функцию координаты точки $M_4$:
$$ z_(max)=z(-2;-1)=(-2)^3+3\cdot (-2)\cdot (-1)^2-15\cdot (-2)-12\cdot (-1)+1=29. $$
The extremum study is completed. It remains only to write down the answer.
Answer:
- $(2;1)$ - minimum point, $z_(min)=-27$;
- $(-2;-1)$ - maximum point, $z_(max)=29$.
Note
In the general case, there is no need to calculate the value of $\Delta$, because we are only interested in the sign, and not in the specific value of this parameter. For example, for the example No. 2 considered above, at the point $M_3(2;1)$ we have $\Delta=36\cdot(2^2-1^2)$. Here it is obvious that $\Delta > 0$ (since both factors $36$ and $(2^2-1^2)$ are positive) and it is possible not to find a specific value of $\Delta$. True, this remark is useless for typical calculations - they require to bring the calculations to a number :)
Example #3
Investigate the function $z=x^4+y^4-2x^2+4xy-2y^2+3$ for an extremum.
We will follow. First, let's find the partial derivatives of the first order:
$$ \frac(\partial z)(\partial x)=4x^3-4x+4y; \frac(\partial z)(\partial y)=4y^3+4x-4y. $$
Compose the system of equations $ \left \( \begin(aligned) & \frac(\partial z)(\partial x)=0;\\ & \frac(\partial z)(\partial y)=0. \end( aligned)\right.$:
$$ \left \( \begin(aligned) & 4x^3-4x+4y=0;\\ & 4y^3+4x-4y=0. \end(aligned) \right. $$
Let's reduce both equations by $4$:
$$ \left \( \begin(aligned) & x^3-x+y=0;\\ & y^3+x-y=0. \end(aligned) \right. $$
Let's add the first equation to the second one and express $y$ in terms of $x$:
$$ y^3+x-y+(x^3-x+y)=0;\\ y^3+x^3=0; y^3=-x^3; y=-x. $$
Substituting $y=-x$ into the first equation of the system, we will have:
$$ x^3-x-x=0;\\ x^3-2x=0;\\ x(x^2-2)=0. $$
From the resulting equation we have: $x=0$ or $x^2-2=0$. It follows from the equation $x^2-2=0$ that $x=-\sqrt(2)$ or $x=\sqrt(2)$. So, three values of $x$ are found, namely: $x_1=0$, $x_2=-\sqrt(2)$, $x_3=\sqrt(2)$. Since $y=-x$, then $y_1=-x_1=0$, $y_2=-x_2=\sqrt(2)$, $y_3=-x_3=-\sqrt(2)$.
The first step of the solution is over. We got three stationary points: $M_1(0;0)$, $M_2(-\sqrt(2),\sqrt(2))$, $M_3(\sqrt(2),-\sqrt(2))$ .
Now let's get down to the algorithm. Let's find partial derivatives of the second order:
$$ \frac(\partial^2 z)(\partial x^2)=12x^2-4; \frac(\partial^2 z)(\partial y^2)=12y^2-4; \frac(\partial^2 z)(\partial x \partial y)=4. $$
Find $\Delta$:
$$ \Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\partial^2z)( \partial x\partial y) \right)^2= (12x^2-4)(12y^2-4)-4^2=\\ =4(3x^2-1)\cdot 4(3y^2 -1)-16=16(3x^2-1)(3y^2-1)-16=16\cdot((3x^2-1)(3y^2-1)-1). $$
Now we will calculate the value of $\Delta$ at each of the previously found stationary points. Let's start from the point $M_1(0;0)$. At this point we have: $\Delta(M_1)=16\cdot((3\cdot 0^2-1)(3\cdot 0^2-1)-1)=16\cdot 0=0$. Since $\Delta(M_1) = 0$, additional research is required, because nothing definite can be said about the presence of an extremum at the considered point. Let us leave this point alone for the time being and move on to other points.
Let's examine the point $M_2(-\sqrt(2),\sqrt(2))$. At this point we get:
\begin(aligned) & \Delta(M_2)=16\cdot((3\cdot (-\sqrt(2))^2-1)(3\cdot (\sqrt(2))^2-1)- 1)=16\cdot 24=384;\\ & \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_2)=12\cdot (-\sqrt(2) )^2-4=24-4=20. \end(aligned)
Since $\Delta(M_2) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_2) > 0$, then according to $M_2(-\ sqrt(2),\sqrt(2))$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_2$ into the given function:
$$ z_(min)=z(-\sqrt(2),\sqrt(2))=(-\sqrt(2))^4+(\sqrt(2))^4-2(-\sqrt( 2))^2+4\cdot (-\sqrt(2))\sqrt(2)-2(\sqrt(2))^2+3=-5. $$
Similarly to the previous point, we examine the point $M_3(\sqrt(2),-\sqrt(2))$. At this point we get:
\begin(aligned) & \Delta(M_3)=16\cdot((3\cdot (\sqrt(2))^2-1)(3\cdot (-\sqrt(2))^2-1)- 1)=16\cdot 24=384;\\ & \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3)=12\cdot (\sqrt(2)) ^2-4=24-4=20. \end(aligned)
Since $\Delta(M_3) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3) > 0$, then according to $M_3(\sqrt (2),-\sqrt(2))$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_3$ into the given function:
$$ z_(min)=z(\sqrt(2),-\sqrt(2))=(\sqrt(2))^4+(-\sqrt(2))^4-2(\sqrt(2 ))^2+4\cdot \sqrt(2)(-\sqrt(2))-2(-\sqrt(2))^2+3=-5. $$
It's time to return to the point $M_1(0;0)$, where $\Delta(M_1) = 0$. Additional research is required. This evasive phrase means "do what you want" :). There is no general way to resolve such situations - and this is understandable. If there were such a method, then it would have entered all textbooks long ago. In the meantime, we have to look for a special approach to each point at which $\Delta = 0$. Well, let's investigate the behavior of the function in the vicinity of the point $M_1(0;0)$. We note right away that $z(M_1)=z(0;0)=3$. Assume that $M_1(0;0)$ is a minimum point. Then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we get $z(M) > z(M_1) $, i.e. $z(M) > 3$. What if any neighborhood contains points where $z(M)< 3$? Тогда в точке $M_1$ уж точно не будет минимума.
Consider points for which $y=0$, i.e. points of the form $(x,0)$. At these points, the $z$ function will take on the following values:
$$ z(x,0)=x^4+0^4-2x^2+4x\cdot 0-2\cdot 0^2+3=x^4-2x^2+3=x^2(x ^2-2)+3. $$
In all sufficiently small neighborhoods $M_1(0;0)$ we have $x^2-2< 0$, посему $x^2(x^2-2) < 0$, откуда следует $x^2(x^2-2)+3 < 3$. Вывод: любая окрестность точки $M_1(0;0)$ содержит точки, в которых $z < 3$, посему точка $M_1(0;0)$ не может быть точкой минимума.
But maybe the point $M_1(0;0)$ is a maximum point? If this is so, then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we get $z(M)< z(M_1) $, т.е. $z(M) < 3$. А вдруг любая окрестность содержит точки, в которых $z(M) >3$? Then there will definitely not be a maximum at the point $M_1$.
Consider points for which $y=x$, i.e. points of the form $(x,x)$. At these points, the $z$ function will take on the following values:
$$ z(x,x)=x^4+x^4-2x^2+4x\cdot x-2\cdot x^2+3=2x^4+3. $$
Since in any neighborhood of the point $M_1(0;0)$ we have $2x^4 > 0$, then $2x^4+3 > 3$. Conclusion: any neighborhood of the point $M_1(0;0)$ contains points where $z > 3$, so the point $M_1(0;0)$ cannot be a maximum point.
The point $M_1(0;0)$ is neither a maximum nor a minimum. Conclusion: $M_1$ is not an extreme point at all.
Answer: $(-\sqrt(2),\sqrt(2))$, $(\sqrt(2),-\sqrt(2))$ - minimum points of the function $z$. At both points $z_(min)=-5$.
