Find the largest value of a function of several variables. Functions
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Another New Year's Eve... frosty weather and snowflakes on the window glass... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. There is an interesting article on this subject, which contains examples of two-dimensional fractal structures. Here we will look at more complex examples of three-dimensional fractals.
A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, this is a self-similar structure, examining the details of which when magnified, we will see the same shape as without magnification. Whereas in the case of an ordinary geometric figure (not a fractal), upon magnification we will see details that have a simpler shape than the original figure itself. For example, at a high enough magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which will be repeated again and again with each increase.
Benoit Mandelbrot, the founder of the science of fractals, wrote in his article Fractals and Art in the Name of Science: “Fractals are geometric shapes that are as complex in their details as in their overall form. That is, if part of the fractal will be enlarged to the size of the whole, it will appear as a whole, either exactly, or perhaps with a slight deformation."
Theorem 1.5 Let in a closed region D function specified z=z(x,y), having continuous partial derivatives of the first order. Border G region D is piecewise smooth (that is, it consists of pieces of “smooth to the touch” curves or straight lines). Then in the area D function z(x,y) reaches its greatest M and the least m values.
No proof.
You can propose the following plan for finding M And m.
1. We build a drawing, select all parts of the area boundary D and find all the “corner” points of the border.
2. Find stationary points inside D.
3. Find stationary points on each of the boundaries.
4. We calculate at all stationary and corner points, and then select the largest M and least m meanings.
Example 1.14 Find the greatest M and least m function values z= 4x2-2xy+y2-8x in a closed area D, limited: x= 0, y = 0, 4x+3y=12 .
1. Let's build an area D(Fig. 1.5) on a plane Ohoo.
Corner points: O (0; 0), B (0; 4), A (3; 0).
Border G region D consists of three parts:
2. Find stationary points inside the region D:
3. Stationary points on the boundaries l 1, l 2, l 3:
4. We calculate six values:
Examples
Example 1.
This function is defined for all values of the variables x And y, except at the origin, where the denominator goes to zero.
Polynomial x 2 +y 2 is continuous everywhere, and therefore the square root of a continuous function is continuous.
The fraction will be continuous everywhere except at points where the denominator is zero. That is, the function under consideration is continuous on the entire coordinate plane Ohoo, excluding the origin.
Example 2.
Examine the continuity of a function z=tg(x,y). The tangent is defined and continuous for all finite values of the argument, except for values equal to an odd number of the quantity π /2 , i.e. excluding points where
For every fixed "k" equation (1.11) defines a hyperbola. Therefore, the function under consideration is a continuous function xand y, excluding points lying on curves (1.11).
Example 3.
Find partial derivatives of a function u=z -xy, z > 0.
Example 4.
Show that function
satisfies the identity:
– this equality is valid for all points M(x;y;z), except for the point M 0 (a;b;c).
Let's consider the function z=f(x,y) of two independent variables and establish the geometric meaning of the partial variables z"x =f"x(x,y) And z" y =f" y(x,y).
In this case, the equation z=f(x,y) there is an equation of some surface (Fig. 1.3). Let's draw a plane y= const. In a section of this surface plane z=f(x,y) you get some line l 1 intersection along which only the quantities change X And z.
Partial derivative z"x(its geometric meaning directly follows from the known geometric meaning of the derivative of a function of one variable) is numerically equal to the tangent of the angle α tilt, relative to the axis Oh, tangent L 1 to the curve l 1, resulting in a section of the surface z=f(x,y) plane y= const at the point M(x,y,f(xy)): z" x = tanα.
In the section of the surface z=f(x,y) plane X= const you get an intersection line l 2, along which only the quantities change at And z. Then the partial derivative z" y numerically equal to the tangent of the angle β tilt relative to the axis OU, tangent L 2 to the specified line l 2 intersections at a point M(x,y,f(xy)): z" x = tanβ.
Example 5.
What angle does it make with the axis? Oh tangent to line:
at the point M(2,4,5)?
We use the geometric meaning of the partial derivative with respect to a variable X(at constant at):
Example 6.
According to (1.31):
Example 7.
Assuming that the equation
implicitly defines a function
find z"x, z" y.
therefore, according to (1.37), we get the answer.
Example 8.
Explore to the extreme:
1. Find stationary points by solving system (1.41):
that is, four stationary points are found.
2.
by Theorem 1.4 at the point there is a minimum.
Moreover 
4. We calculate six values:
From the six values obtained, select the largest and smallest.
Bibliography:
ü Belko I.V., Kuzmich K.K. Higher mathematics for economists. I semester: Express course. – M.: New knowledge, 2002. – 140 p.
ü Gusak A. A.. Mathematical analysis and differential equations. – Mn.: TetraSystems, 1998. – 416 p.
ü Gusak A. A.. Higher mathematics. A textbook for university students in 2 volumes. – Mn., 1998. – 544 p. (1 volume), 448 pp. (2 volumes).
ü Kremer N. Sh., Putko B. A., Trishin I. M., Fridman M. N. Higher mathematics for economists: Textbook for universities / Ed. prof. N. Sh. Kremer. – M.: UNITI, 2002. – 471 p.
ü Yablonsky A.I., Kuznetsov A.V., Shilkina E.I. and others. Higher mathematics. General course: Textbook / Under general. ed. S. A. Samal. – Mn.: Vysh. school, 2000. – 351 p.
Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let the given function in this region have finite partial derivatives of the first order (except, perhaps, for a finite number of points). To find the largest and smallest values of a function of two variables in a given closed region, three steps of a simple algorithm are required.
Algorithm for finding the largest and smallest values of the function $z=f(x,y)$ in a closed domain $D$.What are critical points? show\hide
Under critical points imply points at which both first-order partial derivatives are equal to zero (i.e. $\frac(\partial z)(\partial x)=0$ and $\frac(\partial z)(\partial y)=0 $) or at least one partial derivative does not exist.
Often the points at which first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.
Example No. 1
Find the largest and smallest values of the function $z=x^2+2xy-y^2-4x$ in a closed region bounded by the lines $x=3$, $y=0$ and $y=x+1$.
We will follow the above, but first we will deal with the drawing of a given area, which we will denote by the letter $D$. We are given the equations of three straight lines that limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the ordinate axis (Oy axis). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct the line $y=x+1$, we will find two points through which we will draw this line. You can, of course, substitute a couple of arbitrary values instead of $x$. For example, substituting $x=10$, we get: $y=x+1=10+1=11$. We have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points at which the straight line $y=x+1$ intersects the lines $x=3$ and $y=0$. Why is this better? Because we will kill a couple of birds with one stone: we will get two points to construct the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines limiting the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ intersects at the point $(-1;0)$. In order not to clutter up the progress of the solution with auxiliary explanations, I will put the question of obtaining these two points in a note.
How were the points $(3;4)$ and $(-1;0)$ obtained? show\hide
Let's start from the intersection point of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second straight lines, therefore, to find the unknown coordinates, you need to solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$
The solution to such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the intersection point of the lines $y=x+1$ and $y=0$. Let us again compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (x-axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything is visible in the picture. However, it is worth remembering that a drawing cannot serve as evidence. The drawing is for illustrative purposes only.
Our area was defined using straight line equations that bound it. Obviously, these lines define a triangle, right? Or is it not entirely obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is incorrect. But to avoid such ambiguities, it is better to define regions by inequalities. Are we interested in the part of the plane located under the straight line $y=x+1$? Ok, so $y ≤ x+1$. Should our area be located above the line $y=0$? Great, that means $y ≥ 0$. By the way, the last two inequalities can easily be combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the region $D$, and they define it unambiguously, without allowing any ambiguity. But how does this help us with the question stated at the beginning of the note? It will also help :) We need to check whether the point $M_1(1;1)$ belongs to the area $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right.$$
Both inequalities are valid. Point $M_1(1;1)$ belongs to region $D$.
Now it’s time to study the behavior of the function at the boundary of the region, i.e. let's go to . Let's start with the straight line $y=0$.
The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Let's substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. We denote the function of one variable $x$ obtained as a result of substitution as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we will also add $M_2(2;0)$ to the list of points. In addition, let us calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points into the original expression $z=x^2+2xy-y^2-4x$. For example, for point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a little. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll write this down in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, there is usually no need for such detailed records, and in the future we will write down all calculations briefly:
$$z_2=f_1(2)=2^2-4\cdot 2=-4;\; z_3=f_1(-1)=(-1)^2-4\cdot (-1)=5;\; z_4=f_1(3)=3^2-4\cdot 3=-3.$$
Now let's turn to the straight line $x=3$. This straight line limits the region $D$ under the condition $0 ≤ y ≤ 4$. Let's substitute $x=3$ into the given function $z$. As a result of this substitution we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$ we need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Let's find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we will also add $M_5(3;3)$ to the previously found points. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at points $M_4(3;0)$ and $M_6(3;4)$. At point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; & z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And finally, consider the last boundary of the region $D$, i.e. straight line $y=x+1$. This straight line limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again we need to find the largest and smallest values of this function on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. Points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We received seven values:
$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$
Let's turn to. Choosing the largest and smallest values from the numbers obtained in the third paragraph, we will have:
$$z_(min)=-4; \; z_(max)=6.$$
The problem is solved, all that remains is to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example No. 2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
First, let's build a drawing. The equation $x^2+y^2=25$ (this is the boundary line of a given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ $25 satisfy all points inside and on the mentioned circle.
We will act according to. Let's find partial derivatives and find out the critical points.
$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. let's find stationary points.
$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8.\end(aligned)\right.$$
We have obtained a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check whether the inequality $x^2+y^2 ≤ 25$ holds, which defines our region $D$. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ does not hold. Conclusion: point $(6;-8)$ does not belong to area $D$.
So, there are no critical points inside the region $D$. Let's move on to... We need to study the behavior of a function on the boundary of a given region, i.e. on the circle $x^2+y^2=25$. We can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the equation of a circle we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ z=x^2+y^2-12x+16y=x^2+25-x^2-12x+16\sqrt(25-x^2)=25-12x+16\sqrt(25-x ^2); \;\; -5≤ x ≤ 5. $$
The further solution will be completely identical to the study of the behavior of the function at the boundary of the region in the previous example No. 1. However, it seems to me more reasonable to apply the Lagrange method in this situation. We will be interested only in the first part of this method. After applying the first part of the Lagrange method, we will obtain points at which we will examine the function $z$ for minimum and maximum values.
We compose the Lagrange function:
$$ F=z(x,y)+\lambda\cdot(x^2+y^2-25)=x^2+y^2-12x+16y+\lambda\cdot (x^2+y^2 -25). $$
We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:
$$ F_(x)^(")=2x-12+2\lambda x; \;\; F_(y)^(")=2y+16+2\lambda y.\\ \left \( \begin (aligned) & 2x-12+2\lambda x=0;\\ & 2y+16+2\lambda y=0;\\ & x^2+y^2-25=0. \end(aligned) \ right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$
To solve this system, let's immediately point out that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ indicates that the value $\lambda=-1$ is unacceptable. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the condition $\lambda\neq -1$. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator $1+\lambda\neq 0$.
Let us substitute the resulting expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \left(\frac(6)(1+\lambda) \right)^2+\left(\frac(-8)(1+\lambda) \right)^2=25;\\ \frac( 36)((1+\lambda)^2)+\frac(64)((1+\lambda)^2)=25;\\ \frac(100)((1+\lambda)^2)=25 ; \; (1+\lambda)^2=4. $$
From the resulting equality it follows that $1+\lambda=2$ or $1+\lambda=-2$. Hence we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we have obtained two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Let's find the values of the function $z$ at points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should select the largest and smallest values from those we obtained in the first and second steps. But in this case the choice is small :) We have:
$$ z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=$125.
From a practical point of view, the greatest interest is in using the derivative to find the largest and smallest values of a function. What is this connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life we have to solve problems of optimizing some parameters. And these are the tasks of finding the largest and smallest values of a function.
It should be noted that the largest and smallest values of a function are usually sought on a certain interval X, which is either the entire domain of the function or part of the domain of definition. The interval X itself can be a segment, an open interval 
, an infinite interval.
In this article we will talk about finding the largest and smallest values of an explicitly defined function of one variable y=f(x) .
Page navigation.
The largest and smallest value of a function - definitions, illustrations.Let's briefly look at the main definitions.
The largest value of the function 
that for anyone 
inequality is true.
The smallest value of the function y=f(x) on the interval X is such a value 
that for anyone inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) accepted value on the interval under consideration at the abscissa.
Stationary points are the values of the argument at which the derivative of the function becomes zero.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. From this theorem it follows that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its largest (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take on its largest and smallest values at points at which the first derivative of this function does not exist, and the function itself is defined.
Let’s immediately answer one of the most common questions on this topic: “Is it always possible to determine the largest (smallest) value of a function”? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of definition of the function, or the interval X is infinite. And some functions at infinity and at the boundaries of the domain of definition can take on both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.
For clarity, we will give a graphic illustration. Look at the pictures and a lot will become clearer.
On the segment
In the first figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside the segment [-6;6].
Consider the case depicted in the second figure. Let's change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest at the point with the abscissa corresponding to the right boundary of the interval.
In Figure 3, the boundary points of the segment [-3;2] are the abscissas of the points corresponding to the largest and smallest value of the function.
On an open interval
In the fourth figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside the open interval (-6;6).
On the interval , no conclusions can be drawn about the largest value.
At infinity
In the example presented in the seventh figure, the function takes the largest value (max y) at a stationary point with abscissa x=1, and the smallest value (min y) is achieved on the right boundary of the interval. At minus infinity, the function values asymptotically approach y=3.
Over the interval, the function reaches neither the smallest nor the largest value. As x=2 approaches from the right, the function values tend to minus infinity (the line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values asymptotically approach y=3. A graphic illustration of this example is shown in Figure 8.
Algorithm for finding the largest and smallest values of a continuous function on a segment.Let us write an algorithm that allows us to find the largest and smallest values of a function on a segment.
Let's analyze the algorithm for solving an example to find the largest and smallest values of a function on a segment.
Example.
Find the largest and smallest value of a function
- on the segment ;
- on the segment [-4;-1] .
Solution.
The domain of definition of a function is the entire set of real numbers, with the exception of zero, that is. Both segments fall within the definition domain.
Find the derivative of the function with respect to: 
Obviously, the derivative of the function exists at all points of the segments and [-4;-1].
We determine stationary points from the equation. The only real root is x=2. This stationary point falls into the first segment.
For the first case, we calculate the values of the function at the ends of the segment and at the stationary point, that is, for x=1, x=2 and x=4: 
Therefore, the greatest value of the function 
is achieved at x=1, and the smallest value 
– at x=2.
For the second case, we calculate the function values only at the ends of the segment [-4;-1] (since it does not contain a single stationary point): 
Solution.
Let's start with the domain of the function. The square trinomial in the denominator of the fraction must not vanish: 
It is easy to check that all intervals from the problem statement belong to the domain of definition of the function.
Let's differentiate the function: 
Obviously, the derivative exists throughout the entire domain of definition of the function.
Let's find stationary points. The derivative goes to zero at . This stationary point falls within the intervals (-3;1] and (-3;2).
Now you can compare the results obtained at each point with the graph of the function. Blue dotted lines indicate asymptotes.
At this point we can finish with finding the largest and smallest values of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest values of the function on any interval. This gives a clearer picture and rigorous justification for the results.
§ Extrema, Maximum and minimum values of functions of several variables - page No. 1/1
§ 8. Extrema. The largest and smallest values of functions of several variables. 1. Extrema of functions of several variables.
plane 
, 
is a point in this area. Dot called maximum point functions 
, if for any point 
inequality holds
.
Likewise point called minimum point functions , if for any point from some neighborhood of a point inequality holds
.
Notes. 1) According to the definitions, the function must be defined in some neighborhood of the point . Those. maximum and minimum points of the function there can only be internal points of the region .
2) If there is a neighborhood of a point , in which for any point different from inequality holds 
(
), then the point called strict maximum point(respectively strict minimum point) functions . In this regard, the maximum and minimum points defined above are sometimes called non-strict maximum and minimum points.
The maximum and minimum points of a function are called its extremum points. The function values at the maximum and minimum points are called respectively highs And minimums, or, in short, extremes this function.
The concepts of extrema are local in nature: the value of a function at a point is compared with the function values at fairly close points. In a given area, a function may have no extrema at all, or it may have several minima, several maxima, and even an infinite number of both. Moreover, some minimums may be greater than some of its maximums. Do not confuse the maximum and minimum values of a function with its maximum and minimum values.
Let us find the necessary condition for an extremum. Let, for example, – maximum point of the function . Then, by definition, there is a gif" align=absmiddle width="17px" height="18px">-neighborhood of the point such that 
for any point 
from this vicinity. In particular,
(1)
Where 
, 
, And
(2)
Where 
, 
. But (1) means that a function of one variable 
has at the point 
maximum or is on the interval 
constant. Hence,
or 
- does not exist,
⇒ 
or 
- does not exist.
Similarly from (2) we obtain that
or 
- does not exist.
Thus, the following theorem is valid.
THEOREM 8.1. (necessary conditions for an extremum). If the function at the point has an extremum, then at this point either both of its first-order partial derivatives are equal to zero, or at least one of these partial derivatives does not exist.
Geometrically, Theorem 8.1 means that if – extremum point of the function , then the tangent plane to the graph of this function at the point is either parallel to the plane , or does not exist at all. To verify this, it is enough to remember how to find the equation of a tangent plane to a surface (see formula (4.6)).
Points satisfying the conditions of Theorem 8.1 are called critical points functions . Just as for a function of one variable, the necessary conditions for an extremum are not sufficient. Those. not every critical point of a function will be its extremum point.
EXAMPLE. Consider the function 
. Dot 
is critical for this function, since at this point both of its first-order partial derivatives 
And 
are equal to zero. However, it will not be an extreme point. Really, 
, but in any neighborhood of the point there are points at which the function takes positive values and points at which the function takes negative values. This is easy to verify if you build a graph of the function - a hyperbolic paraboloid.
For a function of two variables, the most convenient sufficient conditions are given by the following theorem.
THEOREM 8.2. (sufficient conditions for the extremum of a function of two variables). Let – critical point of the function and in some neighborhood of the point the function has continuous partial derivatives up to and including the second order. Let's denote
, 
, 
.
Then 1) if 
, then point 
is not an extremum point;
If we use Theorem 8.2 to investigate the critical point failed (i.e. if 
or the function has no point in the neighborhood at all continuous partial derivatives of the required order), the answer to the question about the presence at a point extremum will give the sign of the function increment at this point.
Indeed, from the definition it follows that if the function has at the point strict maximum then
for all points 
from some neighborhood of a point , or, otherwise 
for all sufficiently small 
And 
. Likewise, if is a point of strict minimum, then for all sufficiently small And inequality will be satisfied 
.
So, to find out whether the critical point is extremum point, it is necessary to examine the increment of the function at this point. If for all small enough And it will preserve the sign, then at the point the function has a strict extremum (minimum if , and the maximum if ).
Comment. The rule remains true for a non-strict extremum, but with the amendment that for some values And the function increment will be zero
EXAMPLE. Find extrema of functions:
1) 
; 2) 
.
1) Function
And 
also exist everywhere. Solving a system of equations 
, 
find two critical points 
And 
.
To study critical points, we apply Theorem 8.2. We have:
, 
, 
.
Let's explore the point :
, 
, 
,
; 
.
Therefore, at the point this function has a minimum, namely 
.
Exploring the critical point :
, 
, 
, 
.
Therefore, the second critical point is not the extremum point of the function.
2) Function
defined everywhere. Its first order partial derivatives 
and they also exist everywhere. Solving a system of equations , find the only critical point .
To study the critical point, we apply Theorem 8.2. We have:
, 
, 
,
, 
, 
, 
.
Determine the presence or absence of an extremum at a point using Theorem 8.2 failed.
Let's examine the sign of the function increment at the point :
If 
, That 
;
If 
, That 
.
Because the 
does not preserve sign in a neighborhood of a point , then at this point the function does not have an extremum.
The definitions of maximum and minimum and the necessary conditions for an extremum are easily transferred to functions of three or more variables. Sufficient conditions for an extremum for a function
(
) variables are not considered in this course due to their complexity. In this case, we will determine the nature of the critical points by the sign of the function increment. 2. The largest and smallest values of a function. Let the function of two variables defined in some area plane 
, 
,
– points of this area. Function value at a point 
called the largest, if for any point from the region inequality holds 
.
Similarly, the value of the function at the point called the smallest, if for any point from the region inequality holds 
.
Earlier, we already said that if a function is continuous and the area – is closed and limited, then the function takes its greatest and smallest values in this area. At the same time, points And can lie both inside the area , and on its border. If the point (or ) lies inside the region , then this will be the maximum (minimum) point of the function , i.e. critical point of a function inside a region . Therefore, to find the largest and smallest values of the function in area need to:
.
