Extremum points of a function and their finding. Increasing, decreasing and extrema of the function. A necessary condition for an extremum of a function of one variable

Passing in these inequalities to the limit as Δ x→ 0 and taking into account that the derivative f "(x 0) exists, and hence the limit on the left does not depend on how Δ x→ 0, we get: for Δ x → 0 – 0 f"(x 0) ≥ 0 and at Δ x → 0 + 0 f"(x 0) ≤ 0. Since f"(x 0) defines a number, then these two inequalities are compatible only if f"(x 0) = 0.

The proved theorem states that the maximum and minimum points can only be among those values ​​of the argument for which the derivative vanishes.

We have considered the case when a function has a derivative at all points of a certain segment. What happens when the derivative does not exist? Consider examples.

y=|x|.

The function does not have a derivative at a point x=0 (at this point, the graph of the function does not have a definite tangent), but at this point the function has a minimum, since y(0)=0, and for all x≠ 0y > 0.

Thus, from the given examples and the formulated theorem it is clear that the function can have an extremum only in two cases: 1) at the points where the derivative exists and is equal to zero; 2) at the point where the derivative does not exist.

However, if at some point x 0 we know that f"(x 0 ) =0, then it cannot be concluded from this that at the point x 0 the function has an extremum.

For example.

But point x=0 is not an extremum point, since to the left of this point the function values ​​are located below the axis Ox, and above on the right.

Values ​​of an argument from the domain of a function, for which the derivative of the function vanishes or does not exist, are called critical points.

It follows from the foregoing that the extremum points of a function are among the critical points, and, however, not every critical point is an extremum point. Therefore, to find the extremum of the function, you need to find all the critical points of the function, and then examine each of these points separately for maximum and minimum. For this, the following theorem serves.

Theorem 2. (A sufficient condition for the existence of an extremum.) Let the function be continuous on some interval containing the critical point x 0 , and is differentiable at all points of this interval (except, perhaps, the point itself x 0). If, when passing from left to right through this point, the derivative changes sign from plus to minus, then at the point x = x 0 the function has a maximum. If, when passing through x 0 from left to right, the derivative changes sign from minus to plus, then the function has a minimum at this point.

Thus, if

f"(x)>0 at x<x 0 and f"(x)< 0 at x > x 0 , then x 0 - maximum point;

Proof. Let us first assume that when passing through x 0, the derivative changes sign from plus to minus, i.e. for all x close to the point x 0 f "(x)> 0 for x< x 0 , f"(x)< 0 for x > x 0 . Let us apply the Lagrange theorem to the difference f(x) - f(x 0 ) = f "(c)(x- x 0), where c lies between x And x 0 .

Let x< x 0 . Then c< x 0 and f "(c)> 0. That's why f "(c)(x-x 0)< 0 and, therefore,

f(x) - f(x 0 )< 0, i.e. f(x)< f(x 0 ).

Let x > x 0 . Then c>x 0 and f"(c)< 0. Means f "(c)(x-x 0)< 0. That's why f(x) - f(x 0 ) <0,т.е.f(x)< f(x 0 ) .

Thus, for all values x close enough to x 0 f(x)< f(x 0 ) . And this means that at the point x 0 the function has a maximum.

The second part of the minimum theorem is proved similarly.

Let us illustrate the meaning of this theorem in the figure. Let f"(x 1 ) =0 and for any x, close enough to x 1 , the inequalities

f"(x)< 0 at x< x 1 , f "(x)> 0 at x > x 1 .

Then to the left of the point x 1 the function is increasing, and decreasing on the right, therefore, when x = x 1 function goes from increasing to decreasing, that is, it has a maximum.

Similarly, one can consider the points x 2 and x 3 .

Schematically, all of the above can be depicted in the picture:

The rule for studying the function y=f(x) for an extremum

Find the scope of a function f(x).

Find the first derivative of a function f"(x).

Determine critical points, for this:

find the real roots of the equation f"(x)=0;

find all values x under which the derivative f"(x) does not exist.

Determine the sign of the derivative to the left and right of the critical point. Since the sign of the derivative remains constant between two critical points, it suffices to determine the sign of the derivative at any one point to the left and at one point to the right of the critical point.

Calculate the value of the function at the extremum points.

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

1. 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.
2. 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:
   1. 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.
   2. 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.
   3. If $\Delta< 0$, то в расматриваемой стационарной точке экстремума нет.
   4. 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 replacement $t=x^2$ (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 $z_(min)=-5$.

A simple algorithm for finding extrema..

• Finding the derivative of a function
• Equate this derivative to zero
• We find the values ​​of the variable of the resulting expression (the values ​​of the variable at which the derivative is converted to zero)
• We divide the coordinate line into intervals with these values ​​(at the same time, we should not forget about the break points, which also need to be applied to the line), all these points are called “suspicious” points for the extremum
• We calculate on which of these intervals the derivative will be positive, and on which it will be negative. To do this, you need to substitute the value from the interval into the derivative.

Of the points suspected of an extremum, it is necessary to find exactly . To do this, we look at our gaps on the coordinate line. If, when passing through some point, the sign of the derivative changes from plus to minus, then this point will be maximum, and if from minus to plus, then minimum.

To find the largest and smallest value of a function, you need to calculate the value of the function at the ends of the segment and at the extremum points. Then choose the largest and smallest value.

Consider an example
We find the derivative and equate it to zero:

We apply the obtained values ​​of the variables to the coordinate line and calculate the sign of the derivative on each of the intervals. Well, for example, for the first take-2 , then the derivative will be-0,24 , for the second take0 , then the derivative will be2 , and for the third we take2 , then the derivative will be-0.24. We put down the appropriate signs.

We see that when passing through the point -1, the derivative changes sign from minus to plus, that is, it will be the minimum point, and when passing through 1, from plus to minus, respectively, this is the maximum point.

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 problem

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; a minimum draft is important for a river ship. 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, the 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 had 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.

With this service, you can find the largest and smallest value of a function one variable f(x) with the design of the solution in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables . You can also find the intervals of increase and decrease of the function.

Function entry rules:

A necessary condition for an extremum of a function of one variable

A sufficient condition for an extremum of a function of one variable

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then the point x * is the point of the local (global) minimum of the function.

If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

That point x * is a local (global) maximum.

Example #1. Find the largest and smallest values ​​of the function: on the segment .
Solution.

The critical point is one x 1 = 2 (f'(x)=0). This point belongs to the segment . (The point x=0 is not critical, since 0∉).
We calculate the values ​​of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 for x=2; f max =9 at x=1

Example #2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let us find the critical points: 1-cos(x)=2, cos(x)=1, x=± π / 3 +2πk, k∈Z. We find y''=2sin(x), calculate , so x= π / 3 +2πk, k∈Z are the minimum points of the function; , so x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example #3. Investigate the extremum function in the neighborhood of the point x=0.
Solution. Here it is necessary to find the extrema of the function. If the extremum x=0 , then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides, the derivative changes sign. At these points, one has to apply other methods to study functions to an extremum.

Example #4. Divide the number 49 into two terms, the product of which will be the largest.
Solution. Let x be the first term. Then (49-x) is the second term.
The product will be maximal: x (49-x) → max