Let's turn to the graph of the function y = x 3 – 3x 2. Let us consider the neighborhood of the point x = 0, i.e. some interval containing this point. It is logical that there is a neighborhood of the point x = 0 such that highest value the function y = x 3 – 3x 2 in this neighborhood takes on the point x = 0. For example, on the interval (-1; 1), the function takes its greatest value equal to 0 at the point x = 0. The point x = 0 is called the maximum point of this functions.

Similarly, the point x = 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 neighborhood of the point x = 2, for example, the neighborhood (1.5; 2.5).

Thus, the maximum point of the function f(x) is called the point x 0 if there is a neighborhood of the point x 0 such that the inequality f(x) ≤ f(x 0) holds for all x from this neighborhood.

For example, point x 0 = 0 is the maximum point of the function f(x) = 1 – x 2, since f(0) = 1 and the inequality f(x) ≤ 1 is true for all values ​​of x.

The minimum point of the function f(x) is a 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, point x 0 = 2 is the minimum point of the function f(x) = 3 + (x – 2) 2, since f(2) = 3 and f(x) ≥ 3 for all x.

Extremum points are called minimum and maximum points.

Let us turn to the function f(x), which is defined in a certain neighborhood of the point x 0 and has a derivative at this point.

If x 0 is the extremum point of the differentiable function f(x), then f "(x 0) = 0. This statement is called Fermat’s theorem.

Fermat's theorem has a visual geometric meaning: at the extremum point the tangent is parallel to the x-axis and therefore its slope
f "(x 0) is equal to zero.

For example, the function f(x) = 1 – 3x2 has a maximum at point x0 = 0, its derivative f "(x) = -2x, f "(0) = 0.

The function f(x) = (x – 2) 2 + 3 has a minimum at point x 0 = 2, f "(x) = 2(x – 2), f "(2) = 0.

Note that if f "(x 0) = 0, then this is not enough to state that x 0 is necessarily the extremum point of the function f (x).

For example, if f(x) = x 3, then f "(0) = 0. However, the point x = 0 is not an extremum point, since the function x 3 increases along the entire numerical axis.

So, the extremum points of the differentiable function must be sought only among the roots of the equation
f "(x) = 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 zero.

Thus, in order for the point x 0 to be an extremum point, it is necessary that it be a stationary point.

Let us consider sufficient conditions for the stationary point to be an extremum point, i.e. conditions under which a stationary point is a point of minimum or maximum of a function.

If the derivative to the left of the stationary point is positive, and to the right – negative, i.e. the derivative changes the “+” sign to the “–” sign when passing through this point, then this stationary point is the maximum point.

Indeed, in 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 the “–” sign to the “+” 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. to the left and right of the stationary point the derivative is positive or negative, then this point is not an extremum point.

Let's consider one of the problems. Find the extremum points of the function f(x) = x 4 – 4x 3.

Solution.

1) Find the derivative: f "(x) = 4x 3 – 12x 2 = 4x 2 (x – 3).

2) Find stationary points: 4x 2 (x – 3) = 0, x 1 = 0, x 2 = 3.

3) Using the interval method, we establish that the derivative f "(x) = 4x 2 (x – 3) is positive for x > 3, negative for x< 0 и при 0 < х < 3.

4) Since when passing through the point x 1 = 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 = 3. Therefore, x 2 = 3 is the minimum point.

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>>Extrema

Extremum of the function

Definition of extremum

Function y = f(x) is called increasing (decreasing) in a certain interval, if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f (x 2)).

If the differentiable function y = f (x) increases (decreases) on an interval, then its derivative on this interval f " (x)> 0

(f"(x)< 0).

Dot x O called local maximum point (minimum) function f (x) if there is a neighborhood of the point x o, for all points of which the inequality f (x) is true≤ f (x o ) (f (x )≥ f (x o )).

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extremes.

Extremum points

The necessary conditions extremum . If the point x O is the extremum point of the function f (x), then either f " (x o ) = 0, or f(x o ) does not exist. Such points are called critical, and the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let x O - critical point. If f" (x ) when passing through a point x O changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If, when passing through the critical point, the derivative does not change sign, then at the point x O there is no extreme.

Second sufficient condition. Let the function f(x) have
f" (x ) in the vicinity of the point x O and the second derivative at the point itself x o. If f"(x o) = 0, >0 ( <0), то точка x o is the local minimum (maximum) point of the function f (x). If =0, then you need to either use the first sufficient condition or involve higher ones.

On a segment, the function y = f (x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22.

Solution. Because f " (

Problems of finding the extremum of a function

Example 3.23. a

Solution. x And y y
0 ≤ x≤
> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions kv. units).

Example 3.24. p ≈

Solution. p p
S"
R = 2, H = 16/4 = 4.

Example 3.22.Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Because f " (x ) = 6x 2 - 30x +36 = 6(x ​​-2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. Since when passing through the point x 1 = 2 the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3, the derivative changes its sign from minus to plus, so at the point x 2 = 3 the function has a minimum. Having calculated the function values ​​at the points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23.It is necessary to build a rectangular area near the stone wall so that it is fenced off on three sides with wire mesh, and the fourth side is adjacent to the wall. For this there is a linear meters of mesh. At what aspect ratio will the site have the largest area?

Solution.Let us denote the sides of the platform by x And y. The area of ​​the site is S = xy. Let y- this is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must be satisfied. Therefore y = a - 2x and S = x (a - 2x), where
0 ≤ x≤ a /2 (the length and width of the area cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2 × a/4 =a/2. Because the x = a /4 is the only critical point; let’s check whether the sign of the derivative changes when passing through this point. At x a /4 S "> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions S(a/4) = a/4(a - a/2) = a 2 /8 (kv. units). Since S is continuous on and its values ​​at the ends S(0) and S(a /2) are equal to zero, then the value found will be the largest value of the function. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24.It is required to manufacture a closed cylindrical tank with a capacity of V=16 p ≈ 50 m 3. What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used for its manufacture?

Solution.The total surface area of ​​the cylinder is S = 2 p R(R+H). We know the volume of the cylinder V = p R 2 Н Þ Н = V/ p R 2 =16 p / p R2 = 16/R2. So S(R) = 2 p (R 2 +16/R). We find the derivative of this function:
S"(R) = 2 p (2R- 16/R 2) = 4 p (R- 8/R 2). S" (R) = 0 at R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

Let the function $z=f(x,y)$ be defined in some neighborhood of the point $(x_0,y_0)$. They say that $(x_0,y_0)$ is a (local) maximum point if for all points $(x,y)$ in some neighborhood of the point $(x_0,y_0)$ the inequality $f(x,y) is satisfied< f(x_0,y_0)$. 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 the (local) minimum point.

The maximum and minimum points are often called the general 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 minimums and maximums of a function are united by a common term - extrema of a function.

Algorithm for studying the function $z=f(x,y)$ for extremum

1. Find the partial derivatives $\frac(\partial z)(\partial x)$ 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 calculate the value of $\Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac (\partial^2z)(\partial x\partial y) \right)^2$ at each 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 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 more complete 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 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 certain quantities is greater than zero, then these quantities are of 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 $\frac(\partial^2z)(\partial y^2)$ coincide.

Example No. 1

Examine the function $z=4x^2-6xy-34x+5y^2+42y+7$ for its 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 move the numbers to the right 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 most convenient to use the Cramer 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 No. 2

Examine the function $z=x^3+3xy^2-15x-12y+1$ for its extremum.

We will follow the above. First, let's find the first-order partial derivatives:

$$ \frac(\partial z)(\partial x)=3x^2+3y^2-15; \frac(\partial z)(\partial y)=6xy-12. $$

Let's create a 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. $$

Let's 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 will 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 $t=x^2$ (meaning 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 started with the algorithm. Let's find the second order partial derivatives:

$$ \frac(\partial^2 z)(\partial x^2)=6x; \frac(\partial^2 z)(\partial y^2)=6x; \frac(\partial^2 z)(\partial x \partial y)=6y. $$

Let's 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 examine 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. All that remains is 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 specific meaning this parameter. For example, for example No. 2 considered above, at 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, for standard calculations this remark is useless - there they require you to bring the calculations to a number :)

Example No. 3

Examine the function $z=x^4+y^4-2x^2+4xy-2y^2+3$ for its extremum.

We will follow. First, let's find the first-order partial derivatives:

$$ \frac(\partial z)(\partial x)=4x^3-4x+4y; \frac(\partial z)(\partial y)=4y^3+4x-4y. $$

Let's create a 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 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 $x^2-2=0$. From the equation $x^2-2=0$ it follows 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 completed. 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 started with the algorithm. Let's find the second order partial derivatives:

$$ \frac(\partial^2 z)(\partial x^2)=12x^2-4; \frac(\partial^2 z)(\partial y^2)=12y^2-4; \frac(\partial^2 z)(\partial x \partial y)=4. $$

Let's 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 $\Delta(M_1) = 0$, then additional research is required, since nothing definite can be said about the presence of an extremum at the point under consideration. Let's leave this point alone for now 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 $M_1(0;0)$, at which $\Delta(M_1) = 0$. According to this, 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 such a method existed, it would have been included in 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 examine the behavior of the function in the vicinity of the point $M_1(0;0)$. Let us immediately note that $z(M_1)=z(0;0)=3$. Let's assume that $M_1(0;0)$ is the minimum point. Then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we obtain $z(M) > z(M_1)$, i.e. $z(M) > 3$. What if any neighborhood contains points at which $z(M)< 3$? Тогда в точке $M_1$ уж точно не будет минимума.

Let us consider points for which $y=0$, i.e. points of the form $(x,0)$. At these points the function $z$ will take 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 the maximum point? If this is so, then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we obtain $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 be no maximum at point $M_1$.

Let's consider points for which $y=x$, i.e. points of the form $(x,x)$. At these points the function $z$ will take 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 at which $z > 3$, therefore the point $M_1(0;0)$ cannot be a maximum point.

Point $M_1(0;0)$ is neither a maximum nor a minimum point. Conclusion: $M_1$ is not an extremum point at all.

Answer: $(-\sqrt(2),\sqrt(2))$, $(\sqrt(2),-\sqrt(2))$ are the minimum points of the function $z$. At both points $z_(min)=-5$.

To determine the nature of a function and talk about its behavior, it is necessary to find intervals of increase and decrease. This process is called function research and graphing. The extremum point is used when finding the largest and smallest values ​​of a function, since at them the function increases or decreases from the interval.

This article reveals the definitions, formulates a sufficient sign of increase and decrease on the interval and a condition for the existence of an extremum. This applies to solving examples and problems. The section on differentiating functions should be repeated, because the solution will need to use finding the derivative.

Yandex.RTB R-A-339285-1 Definition 1

The function y = f (x) will increase on the interval x when, for any x 1 ∈ X and x 2 ∈ X, x 2 > x 1, the inequality f (x 2) > f (x 1) is satisfied. In other words, a larger value of the argument corresponds to a larger value of the function.

Definition 2

The function y = f (x) is considered to be decreasing on the interval x when, for any x 1 ∈ X, x 2 ∈ X, x 2 > x 1, the equality f (x 2) > f (x 1) is considered true. In other words, a larger function value corresponds to a smaller argument value. Consider the figure below.

Comment: When the function is definite and continuous at the ends of the interval of increasing and decreasing, that is (a; b), where x = a, x = b, the points are included in the interval of increasing and decreasing. This does not contradict the definition; it means that it takes place on the interval x.

Basic properties elementary functions type y = sin x – definiteness and continuity for real values ​​of the arguments. From here we get that the sine increases over the interval - π 2; π 2, then the increase on the segment has the form - π 2; π 2.

The point x 0 is called maximum point for the function y = f (x), when for all values ​​of x the inequality f (x 0) ≥ f (x) is valid. Maximum function is the value of the function at a point, and is denoted by y m a x .

The point x 0 is called the minimum point for the function y = f (x), when for all values ​​of x the inequality f (x 0) ≤ f (x) is valid. Minimum functions is the value of the function at a point, and has a designation of the form y m i n .

Neighborhoods of the point x 0 are considered extremum points, and the value of the function that corresponds to the extremum points. Consider the figure below.

Extrema of a function with the largest and smallest value of the function. Consider the figure below.

The first figure says that it is necessary to find the largest value of the function from the segment [a; b ] . It is found using maximum points and is equal to the maximum value of the function, and the second figure is more like finding the maximum point at x = b.

Sufficient conditions for a function to increase and decrease

To find the maxima and minima of a function, it is necessary to apply signs of extremum in the case when the function satisfies these conditions. The first sign is considered the most frequently used.

The first sufficient condition for an extremum

Let a function y = f (x) be given, which is differentiable in an ε neighborhood of the point x 0, and has continuity at the given point x 0. From here we get that

• when f " (x) > 0 with x ∈ (x 0 - ε ; x 0) and f " (x)< 0 при x ∈ (x 0 ; x 0 + ε) , тогда x 0 является точкой максимума;
• when f "(x)< 0 с x ∈ (x 0 - ε ; x 0) и f " (x) >0 for x ∈ (x 0 ; x 0 + ε), then x 0 is the minimum point.

In other words, we obtain their conditions for setting the sign:

• when the function is continuous at the point x 0, then it has a derivative with a changing sign, that is, from + to -, which means the point is called a maximum;
• when the function is continuous at the point x 0, then it has a derivative with a changing sign from - to +, which means the point is called a minimum.

To correctly determine the maximum and minimum points of a function, you must follow the algorithm for finding them:

• find the domain of definition;
• find the derivative of the function on this area;
• identify zeros and points where the function does not exist;
• determining the sign of the derivative on intervals;
• select points where the function changes sign.

Let's consider the algorithm by solving several examples of finding extrema of a function.

Example 1

Find the maximum and minimum points of the given function y = 2 (x + 1) 2 x - 2 .

Solution

The domain of definition of this function is all real numbers except x = 2. First, let's find the derivative of the function and get:

y " = 2 x + 1 2 x - 2 " = 2 x + 1 2 " (x - 2) - (x + 1) 2 (x - 2) " (x - 2) 2 = = 2 2 (x + 1) (x + 1) " (x - 2) - (x + 1) 2 1 (x - 2) 2 = 2 2 (x + 1) (x - 2 ) - (x + 2) 2 (x - 2) 2 = = 2 · (x + 1) · (x - 5) (x - 2) 2

From here we see that the zeros of the function are x = - 1, x = 5, x = 2, that is, each bracket must be equated to zero. Let's mark it on the number axis and get:

Now we determine the signs of the derivative from each interval. It is necessary to select a point included in the interval and substitute it into the expression. For example, points x = - 2, x = 0, x = 3, x = 6.

We get that

y " (- 2) = 2 · (x + 1) · (x - 5) (x - 2) 2 x = - 2 = 2 · (- 2 + 1) · (- 2 - 5) (- 2 - 2) 2 = 2 · 7 16 = 7 8 > 0, which means that the interval - ∞ ; - 1 has a positive derivative. Similarly, we find that

y " (0) = 2 · (0 + 1) · 0 - 5 0 - 2 2 = 2 · - 5 4 = - 5 2< 0 y " (3) = 2 · (3 + 1) · (3 - 5) (3 - 2) 2 = 2 · - 8 1 = - 16 < 0 y " (6) = 2 · (6 + 1) · (6 - 5) (6 - 2) 2 = 2 · 7 16 = 7 8 > 0

Since the second interval turned out to be less than zero, it means that the derivative on the interval will be negative. The third with a minus, the fourth with a plus. To determine continuity, you need to pay attention to the sign of the derivative; if it changes, then this is an extremum point.

We find that at the point x = - 1 the function will be continuous, which means that the derivative will change sign from + to -. According to the first sign, we have that x = - 1 is a maximum point, which means we get

y m a x = y (- 1) = 2 (x + 1) 2 x - 2 x = - 1 = 2 (- 1 + 1) 2 - 1 - 2 = 0

The point x = 5 indicates that the function is continuous, and the derivative will change sign from – to +. This means that x = -1 is the minimum point, and its determination has the form

y m i n = y (5) = 2 (x + 1) 2 x - 2 x = 5 = 2 (5 + 1) 2 5 - 2 = 24

Graphic image

Answer: y m a x = y (- 1) = 0, y m i n = y (5) = 24.

It is worth paying attention to the fact that the use of the first sufficient criterion for an extremum does not require the differentiability of the function at the point x 0, this simplifies the calculation.

Example 2

Find the maximum and minimum points of the function y = 1 6 x 3 = 2 x 2 + 22 3 x - 8.

Solution.

The domain of a function is all real numbers. This can be written as a system of equations of the form:

1 6 x 3 - 2 x 2 - 22 3 x - 8 , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 , x ≥ 0

Then you need to find the derivative:

y " = 1 6 x 3 - 2 x 2 - 22 3 x - 8 " , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 " , x >0 y " = - 1 2 x 2 - 4 x - 22 3 , x< 0 1 2 x 2 - 4 x + 22 3 , x > 0

The point x = 0 does not have a derivative, because the values ​​of the one-sided limits are different. We get that:

lim y "x → 0 - 0 = lim y x → 0 - 0 - 1 2 x 2 - 4 x - 22 3 = - 1 2 (0 - 0) 2 - 4 (0 - 0) - 22 3 = - 22 3 lim y " x → 0 + 0 = lim y x → 0 - 0 1 2 x 2 - 4 x + 22 3 = 1 2 (0 + 0) 2 - 4 (0 + 0) + 22 3 = + 22 3

It follows that the function is continuous at the point x = 0, then we calculate

lim y x → 0 - 0 = lim x → 0 - 0 - 1 6 x 3 - 2 x 2 - 22 3 x - 8 = = - 1 6 · (0 - 0) 3 - 2 · (0 - 0) 2 - 22 3 (0 - 0) - 8 = - 8 lim y x → 0 + 0 = lim x → 0 - 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 = = 1 6 (0 + 0) 3 - 2 · (0 + 0) 2 + 22 3 · (0 + 0) - 8 = - 8 y (0) = 1 6 x 3 - 2 x 2 + 22 3 x - 8 x = 0 = 1 6 · 0 3 - 2 0 2 + 22 3 0 - 8 = - 8

It is necessary to make calculations to find the value of the argument when the derivative becomes equal to zero:

1 2 x 2 - 4 x - 22 3 , x< 0 D = (- 4) 2 - 4 · - 1 2 · - 22 3 = 4 3 x 1 = 4 + 4 3 2 · - 1 2 = - 4 - 2 3 3 < 0 x 2 = 4 - 4 3 2 · - 1 2 = - 4 + 2 3 3 < 0

1 2 x 2 - 4 x + 22 3 , x > 0 D = (- 4) 2 - 4 1 2 22 3 = 4 3 x 3 = 4 + 4 3 2 1 2 = 4 + 2 3 3 > 0 x 4 = 4 - 4 3 2 1 2 = 4 - 2 3 3 > 0

All obtained points must be marked on a straight line to determine the sign of each interval. Therefore, it is necessary to calculate the derivative at arbitrary points for each interval. For example, we can take points with values ​​x = - 6, x = - 4, x = - 1, x = 1, x = 4, x = 6. We get that

y " (- 6) = - 1 2 x 2 - 4 x - 22 3 x = - 6 = - 1 2 · - 6 2 - 4 · (- 6) - 22 3 = - 4 3< 0 y " (- 4) = - 1 2 x 2 - 4 x - 22 3 x = - 4 = - 1 2 · (- 4) 2 - 4 · (- 4) - 22 3 = 2 3 >0 y " (- 1) = - 1 2 x 2 - 4 x - 22 3 x = - 1 = - 1 2 · (- 1) 2 - 4 · (- 1) - 22 3 = 23 6< 0 y " (1) = 1 2 x 2 - 4 x + 22 3 x = 1 = 1 2 · 1 2 - 4 · 1 + 22 3 = 23 6 >0 y "(4) = 1 2 x 2 - 4 x + 22 3 x = 4 = 1 2 4 2 - 4 4 + 22 3 = - 2 3< 0 y " (6) = 1 2 x 2 - 4 x + 22 3 x = 6 = 1 2 · 6 2 - 4 · 6 + 22 3 = 4 3 > 0

The image on the straight line looks like

This means that we come to the conclusion that it is necessary to resort to the first sign of an extremum. Let us calculate and find that

x = - 4 - 2 3 3 , x = 0 , x = 4 + 2 3 3 , then from here the maximum points have the values ​​x = - 4 + 2 3 3 , x = 4 - 2 3 3

Let's move on to calculating the minimums:

y m i n = y - 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 0 = - 8 y m i n = y 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 + 2 3 3 = - 8 27 3

Let's calculate the maxima of the function. We get that

y m a x = y - 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 - 2 3 3 = 8 27 3

Graphic image

Answer:

y m i n = y - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = - 8 y m i n = y 4 + 2 3 3 = - 8 27 3 y m a x = y - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 8 27 3

If a function f " (x 0) = 0 is given, then if f "" (x 0) > 0, we obtain that x 0 is a minimum point if f "" (x 0)< 0 , то точкой максимума. Признак связан с нахождением производной в точке x 0 .

Example 3

Find the maxima and minima of the function y = 8 x x + 1.

Solution

First, we find the domain of definition. We get that

D(y) : x ≥ 0 x ≠ - 1 ⇔ x ≥ 0

It is necessary to differentiate the function, after which we get

y " = 8 x x + 1 " = 8 x " (x + 1) - x (x + 1) " (x + 1) 2 = = 8 1 2 x (x + 1) - x 1 (x + 1) 2 = 4 x + 1 - 2 x (x + 1) 2 x = 4 - x + 1 (x + 1) 2 x

At x = 1, the derivative becomes zero, which means that the point is a possible extremum. To clarify, it is necessary to find the second derivative and calculate the value at x = 1. We get:

y "" = 4 - x + 1 (x + 1) 2 x " = = 4 (- x + 1) " (x + 1) 2 x - (- x + 1) x + 1 2 x " (x + 1) 4 x = = 4 (- 1) (x + 1) 2 x - (- x + 1) x + 1 2 " x + (x + 1) 2 x " (x + 1) 4 x = = 4 - (x + 1) 2 x - (- x + 1) 2 x + 1 (x + 1) " x + (x + 1) 2 2 x (x + 1) 4 x = = - (x + 1) 2 x - (- x + 1) x + 1 2 x + x + 1 2 x (x + 1) 4 x = = 2 3 x 2 - 6 x - 1 x + 1 3 x 3 ⇒ y "" (1) = 2 3 1 2 - 6 1 - 1 (1 + 1) 3 (1) 3 = 2 · - 4 8 = - 1< 0

This means that using the 2 sufficient condition for an extremum, we obtain that x = 1 is a maximum point. Otherwise, the entry looks like y m a x = y (1) = 8 1 1 + 1 = 4.

Graphic image

Answer: y m a x = y (1) = 4 ..

The function y = f (x) has its derivative up to the nth order in the ε neighborhood of a given point x 0 and its derivative up to the n + 1st order at the point x 0 . Then f " (x 0) = f "" (x 0) = f " " " (x 0) = . . . = f n (x 0) = 0 .

It follows that when n is an even number, then x 0 is considered an inflection point, when n is an odd number, then x 0 is an extremum point, and f (n + 1) (x 0) > 0, then x 0 is a minimum point, f (n + 1) (x 0)< 0 , тогда x 0 является точкой максимума.

Example 4

Find the maximum and minimum points of the function y y = 1 16 (x + 1) 3 (x - 3) 4.

Solution

The original function is a rational entire function, which means that the domain of definition is all real numbers. It is necessary to differentiate the function. We get that

y " = 1 16 x + 1 3 " (x - 3) 4 + (x + 1) 3 x - 3 4 " = = 1 16 (3 (x + 1) 2 (x - 3) 4 + (x + 1) 3 4 (x - 3) 3) = = 1 16 (x + 1) 2 (x - 3) 3 (3 x - 9 + 4 x + 4) = 1 16 (x + 1) 2 (x - 3) 3 (7 x - 5)

This derivative will go to zero at x 1 = - 1, x 2 = 5 7, x 3 = 3. That is, the points can be possible extremum points. It is necessary to apply the third sufficient condition for the extremum. Finding the second derivative allows you to accurately determine the presence of a maximum and minimum of a function. The second derivative is calculated at the points of its possible extremum. We get that

y "" = 1 16 x + 1 2 (x - 3) 3 (7 x - 5) " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) y "" (- 1) = 0 y "" 5 7 = - 36864 2401< 0 y "" (3) = 0

This means that x 2 = 5 7 is the maximum point. Applying the 3rd sufficient criterion, we obtain that for n = 1 and f (n + 1) 5 7< 0 .

It is necessary to determine the nature of the points x 1 = - 1, x 3 = 3. To do this, you need to find the third derivative and calculate the values ​​at these points. We get that

y " " " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) " = = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) y " " " (- 1) = 96 ≠ 0 y " " " (3) = 0

This means that x 1 = - 1 is the inflection point of the function, since for n = 2 and f (n + 1) (- 1) ≠ 0. It is necessary to investigate the point x 3 = 3. To do this, we find the 4th derivative and perform calculations at this point:

y (4) = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) " = = 1 2 (105 x 3 - 405 x 2 + 315 x + 57) y (4) ( 3) = 96 > 0

From what was decided above we conclude that x 3 = 3 is the minimum point of the function.

Graphic image

Answer: x 2 = 5 7 is the maximum point, x 3 = 3 is the minimum point of the given function.

If you notice an error in the text, please highlight it and press Ctrl+Enter

Find the largest value of the function y=(7x^2-56x+56)e^x on the segment [-3; 2].

Solution

Let's find the derivative of the original function using the product derivative formula y"=(7x^2-56x+56)"e^x\,+ (7x^2-56x+56)\left(e^x\right)"= (14x-56)e^x+(7x^2-56x+56)e^x= (7x^2-42x)e^x= 7x(x-6)e^x. Let's calculate the zeros of the derivative: y"=0;

7x(x-6)e^x=0,

x_1=0, x_2=6.

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function on a given segment.

From the figure it is clear that on the segment [-3; 0] the original function increases, and on the segment it decreases. Thus, the largest value on the segment [-3; 2] is achieved at x=0 and is equal to y(0)= 7\cdot 0^2-56\cdot 0+56=56.

Answer

Condition

Find the greatest value of the function y=12x-12tg x-18 on the segment \left.

Solution

y"= (12x)"-12(tg x)"-(18)"= 12-\frac(12)(\cos ^2x)= \frac(12\cos ^2x-12)(\cos ^2x)\leqslant0. This means that the original function is non-increasing on the interval under consideration and takes the greatest value at the left end of the interval, that is, at x=0. The largest value is y(0)= 12\cdot 0-12 tg (0)-18= -18.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level" Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the minimum point of the function y=(x+8)^2e^(x+52).

Solution

We will find the minimum point of the function using the derivative. Let's find the derivative of a given function using the formulas for the derivative of the product, the derivative of x^\alpha and e^x:

y"(x)= \left((x+8)^2\right)"e^(x+52)+(x+8)^2\left(e^(x+52)\right)"= 2(x+8)e^(x+52)+(x+8)^2e^(x+52)= (x+8)e^(x+52)(2+x+8)= (x+8)(x+10)e^(x+52).

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function. e^(x+52)>0 for any x. y"=0 at x=-8, x=-10.

The figure shows that the function y=(x+8)^2e^(x+52) has a single minimum point x=-8.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the maximum point of the function y=8x-\frac23x^\tfrac32-106.

Solution

ODZ: x \geqslant 0. Let's find the derivative of the original function:

y"=8-\frac23\cdot\frac32x^\tfrac12=8-\sqrt x.

Let's calculate the zeros of the derivative:

8-\sqrt x=0;

\sqrt x=8;

x=64.

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function.

The figure shows that point x=64 is the only maximum point of the given function.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find smallest value functions y=5x^2-12x+2\ln x+37 on the segment \left[\frac35; \frac75\right].

Solution

ODZ: x>0.

Let's find the derivative of the original function:

y"(x)= 10x-12+\frac(2)(x)= \frac(10x^2-12x+2)(x).

Let's define the zeros of the derivative: y"(x)=0;

\frac(10x^2-12x+2)(x)=0,

5x^2-6x+1=0,

x_(1,2)= \frac(3\pm\sqrt(3^2-5\cdot1))(5)= \frac(3\pm2)(5),

x_1=\frac15\notin\left[\frac35; \frac75\right],

x_2=1\in\left[\frac35; \frac75\right].

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function on the interval under consideration.

From the figure it is clear that on the segment \left[\frac35; 1\right] the original function decreases, and on the segment \left increases. Thus, the smallest value on the segment \left[\frac35; \frac75\right] is achieved at x=1 and is equal to y(1)= 5\cdot 1^2-12\cdot 1+2 \ln 1+37= 30.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the greatest value of the function y=(x+4)^2(x+1)+19 on the segment [-5; -3].

Solution

Let's find the derivative of the original function using the product derivative formula.