Extremum points and their characteristics. Increasing and decreasing functions on the interval, extremums. We continue to search for extremums of the function together

The point x 0 is called maximum point(minimum) of the function f(х) if in some neighborhood of the point x 0 the inequality f(х) ≤f(х 0) (f(х) ≥f(х 0)) is fulfilled.

The value of the function at this point is called accordingly maximum or minimum functions. The maximum and minimum of a function are combined by a common name extremum functions.

The extremum of a function in this sense is often called local extremum, emphasizing the fact that this concept is associated only with a sufficiently small neighborhood of the point x 0 . On the same interval, a function can have several local maxima and minima, which do not necessarily coincide with global maximum or minimum(i.e. the largest or smallest value of the function on the entire interval).

Necessary condition for an extremum. In order for a function to have an extremum at a point, it is necessary that its derivative at that point be equal to zero or not exist.

For differentiable functions, this condition follows from Fermat's theorem. In addition, it provides for the case when the function has an extremum at a point where it is not differentiable.

The points at which the necessary extremum condition is satisfied are called critical(or stationary for a differentiable function). These points must be within the scope of the function.

Thus, if there is an extremum at any point, then this point is critical (need condition). Note that the converse is not true. The critical point is not necessarily an extremum point, i.e. the stated condition is not sufficient.

The first sufficient condition for an extremum. If, when passing through a certain point, the derivative of a differentiable function changes its sign from plus to minus, then this is the maximum point of the function, and if from minus to plus, then the minimum point.

The proof of this condition follows from the sufficient condition of monotonicity (when the sign of the derivative changes, the transition occurs either from an increase in the function to a decrease, or from a decrease to an increase).

The second sufficient condition for an extremum. If the first derivative of a twice differentiable function is zero at some point, and the second derivative is positive at that point, then this is the minimum point of the function; and if the second derivative is negative, then this is the maximum point.

The proof of this condition is also based on the sufficient monotonicity condition. Indeed, if the second derivative is positive, then the first derivative is an increasing function. Since it is equal to zero at the point under consideration, therefore, when passing through it, it changes sign from minus to plus, which returns us to the first sufficient condition for a local minimum. Similarly, if the second derivative is negative, then the first one decreases and changes sign from plus to minus, which is a sufficient condition for a local maximum.

Investigation of a function to an extremum in accordance with the formulated theorems, it includes the following steps:

1. Find the first derivative of the function f`(x).

2. Check the fulfillment of the necessary extremum condition, i.e. find critical points of the function f(x) at which the derivative f`(x) = 0 or does not exist.

3. Check the fulfillment of the sufficient extremum condition, i.e. either examine the sign of the derivative to the left and right of each critical point, or find the second derivative f``(x) and determine its sign at each critical point. Make a conclusion about the presence of extremums of the function.

4. Find extrema (extreme values) of the function.

Finding the global maximum and minimum of a function on a certain interval is also of great practical importance. The solution of this problem on a segment is based on the Weierstrass theorem, according to which a continuous function takes its largest and smallest values ​​on a segment. They can be achieved both at extremum points and at the ends of the segment. Therefore, the solution includes the following steps:

1. Find the derivative of the function f`(x).

2. Find the critical points of the function f(x) at which the derivative f`(x) = 0 or does not exist.

3. Find the values ​​of the function at critical points and at the ends of the segment and choose the largest and smallest of them.

It can also be said that at these points the direction of movement of the function changes: if the function stops falling and starts growing, this is a minimum point, on the contrary, a maximum.

The minimums and maximums are collectively referred to as function extrema.

In other words, all five points highlighted on the chart above are extreme points.

Thanks to this, finding these points is not a problem, even if you do not have a graph of the function.

Attention! When they write extremes or highs/lows mean the value of the function i.e. \(y\). When they write extreme points or Highs/Lows refers to the X's at which Highs/Lows are reached. For example, in the figure above, \(-5\) is the minimum (or extremum) point, and \(1\) is the minimum (or extremum).

How to find the extremum points of a function according to the graph of the derivative (7 task of the exam)?

Let's find together the number of extremum points of the function according to the graph of the derivative using an example:

We have a graph - so we are looking for at what points on the graph the derivative is equal to zero. Obviously, these are the points \(-13\), \(-11\), \(-9\),\(-7\) and \(3\). The number of extremum points of the function is \(5\).

Attention! If given a schedule derivative functions, but you need to find extremum points of the function, we do not count the highs and lows of the derivative! We count the points at which the derivative of the function vanishes (i.e. crosses the \(x\) axis).

How to find the points of maxima or minima of a function according to the graph of the derivative (7 task of the exam)?

To answer this question, you need to remember two more important rules:

- The derivative is positive where the function increases.
- The derivative is negative where the function decreases.

Using these rules, let's find the minimum and maximum points of the function on the graph of the derivative.

It is clear that the minimums and maximums must be sought among the extremum points, i.e. among \(-13\), \(-11\), \(-9\), \(-7\) and \(3\).

To make it easier to solve the problem, we first place the plus and minus signs in the figure, denoting the sign of the derivative. Then the arrows - denoting the increase, decrease of the function.

Let's start with \(-13\): up to \(-13\) the derivative is positive, i.e. the function grows, after - the derivative is negative i.e. function falls. If you imagine this, it becomes clear that \(-13\) is the maximum point.

\(-11\): the derivative is first positive and then negative, so the function increases and then decreases. Again, try to mentally draw this and it will become obvious to you that \(-11\) is the minimum.

\(- 9\): the function increases and then decreases - the maximum.

\(-7\): minimum.

\(3\): maximum.

All of the above can be summarized in the following conclusions:

- The function has a maximum where the derivative is zero and changes sign from plus to minus.
- The function has a minimum where the derivative is zero and changes sign from minus to plus.

How to find the points of maxima and minima if the formula of the function is known (12 USE task)?

To answer this question, you need to do everything the same as in the previous paragraph: find where the derivative is positive, where it is negative and where it is equal to zero. To make it clearer, I will write an algorithm with an example solution:

1. Find the derivative of the function \(f"(x)\).
2. Find the roots of the equation \(f"(x)=0\).
3. Draw the \(x\) axis and mark the points obtained in step 2 on it, draw the arcs into the intervals into which the axis is divided. Sign above the axis \ (f "(x) \), and under the axis \ (f (x) \).
4. Determine the sign of the derivative in each interval (interval method).
5. Put a sign of the derivative in each gap (above the axis), and use an arrow to indicate the increase (↗) or decrease (↘) of the function (below the axis).
6. Determine how the sign of the derivative has changed when passing through the points obtained in step 2:
   - if \(f'(x)\) changed sign from "\(+\)" to "\(-\)", then \(x_1\) is the maximum point;
   - if \(f'(x)\) changed sign from "\(-\)" to "\(+\)", then \(x_3\) is the minimum point;
   - if \(f'(x)\) has not changed sign, then \(x_2\) can be an inflection point.

Everything! High and low points found.

Depicting points on the axis at which the derivative is equal to zero, the scale can be ignored. The behavior of the function can be shown as shown in the figure below. So it will be more obvious where is the maximum and where is the minimum.

Example(USE). Find the maximum point of the function \(y=3x^5-20x^3-54\).
Decision:
1. Find the derivative of the function: \(y"=15x^4-60x^2\).
2. Equate it to zero and solve the equation:

\(15x^4-60x^2=0\) \(|:15\)
\(x^4-4x^2=0\)
\(x^2 (x^2-4)=0\)
\(x=0\) \(x^2-4=0\)
\(x=±2\)

3. - 6. Let's put points on the real axis and determine how the sign of the derivative changes and how the function moves:

Now it is obvious that the maximum point is \(-2\).

Answer. \(-2\).

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 substitution $t=x^2$ (we keep in mind that $t > 0$):

$$ t^2-5t+4=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 1 \cdot 4=9;\\ & t_1=\frac(-(- 5)-\sqrt(9))(2)=\frac(5-3)(2)=1;\\ & t_2=\frac(-(-5)+\sqrt(9))(2)= \frac(5+3)(2)=4.\end(aligned) $$

If $t=1$, then $x^2=1$. Hence we have two values ​​of $x$: $x_1=1$, $x_2=-1$. If $t=4$, then $x^2=4$, i.e. $x_3=2$, $x_4=-2$. Remembering that $y=\frac(2)(x)$, we get:

\begin(aligned) & y_1=\frac(2)(x_1)=\frac(2)(1)=2;\\ & y_2=\frac(2)(x_2)=\frac(2)(-1 )=-2;\\ & y_3=\frac(2)(x_3)=\frac(2)(2)=1;\\ & y_4=\frac(2)(x_4)=\frac(2)( -2)=-1. \end(aligned)

So, we have four stationary points: $M_1(1;2)$, $M_2(-1;-2)$, $M_3(2;1)$, $M_4(-2;-1)$. This completes the first step of the algorithm.

Now let's get down to the algorithm. Let's find partial derivatives of the second order:

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

Find $\Delta$:

$$ \Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\partial^2z)( \partial x\partial y) \right)^2= 6x\cdot 6x-(6y)^2=36x^2-36y^2=36(x^2-y^2). $$

Now we will calculate the value of $\Delta$ at each of the previously found stationary points. Let's start from the point $M_1(1;2)$. At this point we have: $\Delta(M_1)=36(1^2-2^2)=-108$. Since $\Delta(M_1)< 0$, то согласно в точке $M_1$ экстремума нет.

Let's explore the point $M_2(-1;-2)$. At this point we have: $\Delta(M_2)=36((-1)^2-(-2)^2)=-108$. Since $\Delta(M_2)< 0$, то согласно в точке $M_2$ экстремума нет.

Let's examine the point $M_3(2;1)$. At this point we get:

$$ \Delta(M_3)=36(2^2-1^2)=108;\;\; \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3)=6\cdot 2=12. $$

Since $\Delta(M_3) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3) > 0$, then according to $M_3(2; 1)$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_3$ into the given function:

$$ z_(min)=z(2;1)=2^3+3\cdot 2\cdot 1^2-15\cdot 2-12\cdot 1+1=-27. $$

It remains to explore the point $M_4(-2;-1)$. At this point we get:

$$ \Delta(M_4)=36((-2)^2-(-1)^2)=108;\;\; \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_4)=6\cdot (-2)=-12. $$

Since $\Delta(M_4) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_4)< 0$, то согласно $M_4(-2;-1)$ есть точкой максимума функции $z$. Максимум функции $z$ найдём, подставив в заданную функцию координаты точки $M_4$:

$$ z_(max)=z(-2;-1)=(-2)^3+3\cdot (-2)\cdot (-1)^2-15\cdot (-2)-12\cdot (-1)+1=29. $$

The extremum study is completed. It remains only to write down the answer.

Answer:

• $(2;1)$ - minimum point, $z_(min)=-27$;
• $(-2;-1)$ - maximum point, $z_(max)=29$.

Note

In the general case, there is no need to calculate the value of $\Delta$, because we are only interested in the sign, and not in the specific value of this parameter. For example, for the example No. 2 considered above, at the point $M_3(2;1)$ we have $\Delta=36\cdot(2^2-1^2)$. Here it is obvious that $\Delta > 0$ (since both factors $36$ and $(2^2-1^2)$ are positive) and it is possible not to find a specific value of $\Delta$. True, this remark is useless for typical calculations - they require to bring the calculations to a number :)

Example #3

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

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

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

Compose the system of equations $ \left \( \begin(aligned) & \frac(\partial z)(\partial x)=0;\\ & \frac(\partial z)(\partial y)=0. \end( aligned)\right.$:

$$ \left \( \begin(aligned) & 4x^3-4x+4y=0;\\ & 4y^3+4x-4y=0. \end(aligned) \right. $$

Let's reduce both equations by $4$:

$$ \left \( \begin(aligned) & x^3-x+y=0;\\ & y^3+x-y=0. \end(aligned) \right. $$

Let's add the first equation to the second one and express $y$ in terms of $x$:

$$ y^3+x-y+(x^3-x+y)=0;\\ y^3+x^3=0; y^3=-x^3; y=-x. $$

Substituting $y=-x$ into the first equation of the system, we will have:

$$ x^3-x-x=0;\\ x^3-2x=0;\\ x(x^2-2)=0. $$

From the resulting equation we have: $x=0$ or $x^2-2=0$. It follows from the equation $x^2-2=0$ that $x=-\sqrt(2)$ or $x=\sqrt(2)$. So, three values ​​of $x$ are found, namely: $x_1=0$, $x_2=-\sqrt(2)$, $x_3=\sqrt(2)$. Since $y=-x$, then $y_1=-x_1=0$, $y_2=-x_2=\sqrt(2)$, $y_3=-x_3=-\sqrt(2)$.

The first step of the solution is over. We got three stationary points: $M_1(0;0)$, $M_2(-\sqrt(2),\sqrt(2))$, $M_3(\sqrt(2),-\sqrt(2))$ .

Now let's get down to the algorithm. Let's find partial derivatives of the second order:

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

Find $\Delta$:

$$ \Delta=\frac(\partial^2z)(\partial x^2)\cdot \frac(\partial^2z)(\partial y^2)-\left(\frac(\partial^2z)( \partial x\partial y) \right)^2= (12x^2-4)(12y^2-4)-4^2=\\ =4(3x^2-1)\cdot 4(3y^2 -1)-16=16(3x^2-1)(3y^2-1)-16=16\cdot((3x^2-1)(3y^2-1)-1). $$

Now we will calculate the value of $\Delta$ at each of the previously found stationary points. Let's start from the point $M_1(0;0)$. At this point we have: $\Delta(M_1)=16\cdot((3\cdot 0^2-1)(3\cdot 0^2-1)-1)=16\cdot 0=0$. Since $\Delta(M_1) = 0$, additional research is required, because nothing definite can be said about the presence of an extremum at the considered point. Let us leave this point alone for the time being and move on to other points.

Let's examine the point $M_2(-\sqrt(2),\sqrt(2))$. At this point we get:

\begin(aligned) & \Delta(M_2)=16\cdot((3\cdot (-\sqrt(2))^2-1)(3\cdot (\sqrt(2))^2-1)- 1)=16\cdot 24=384;\\ & \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_2)=12\cdot (-\sqrt(2) )^2-4=24-4=20. \end(aligned)

Since $\Delta(M_2) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_2) > 0$, then according to $M_2(-\ sqrt(2),\sqrt(2))$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_2$ into the given function:

$$ z_(min)=z(-\sqrt(2),\sqrt(2))=(-\sqrt(2))^4+(\sqrt(2))^4-2(-\sqrt( 2))^2+4\cdot (-\sqrt(2))\sqrt(2)-2(\sqrt(2))^2+3=-5. $$

Similarly to the previous point, we examine the point $M_3(\sqrt(2),-\sqrt(2))$. At this point we get:

\begin(aligned) & \Delta(M_3)=16\cdot((3\cdot (\sqrt(2))^2-1)(3\cdot (-\sqrt(2))^2-1)- 1)=16\cdot 24=384;\\ & \left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3)=12\cdot (\sqrt(2)) ^2-4=24-4=20. \end(aligned)

Since $\Delta(M_3) > 0$ and $\left.\frac(\partial^2 z)(\partial x^2)\right|_(M_3) > 0$, then according to $M_3(\sqrt (2),-\sqrt(2))$ is the minimum point of the function $z$. We find the minimum of the function $z$ by substituting the coordinates of the point $M_3$ into the given function:

$$ z_(min)=z(\sqrt(2),-\sqrt(2))=(\sqrt(2))^4+(-\sqrt(2))^4-2(\sqrt(2 ))^2+4\cdot \sqrt(2)(-\sqrt(2))-2(-\sqrt(2))^2+3=-5. $$

It's time to return to the point $M_1(0;0)$, where $\Delta(M_1) = 0$. Additional research is required. This evasive phrase means "do what you want" :). There is no general way to resolve such situations - and this is understandable. If there were such a method, then it would have entered all textbooks long ago. In the meantime, we have to look for a special approach to each point at which $\Delta = 0$. Well, let's investigate the behavior of the function in the vicinity of the point $M_1(0;0)$. We note right away that $z(M_1)=z(0;0)=3$. Assume that $M_1(0;0)$ is a minimum point. Then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we get $z(M) > z(M_1) $, i.e. $z(M) > 3$. What if any neighborhood contains points where $z(M)< 3$? Тогда в точке $M_1$ уж точно не будет минимума.

Consider points for which $y=0$, i.e. points of the form $(x,0)$. At these points, the $z$ function will take on the following values:

$$ z(x,0)=x^4+0^4-2x^2+4x\cdot 0-2\cdot 0^2+3=x^4-2x^2+3=x^2(x ^2-2)+3. $$

In all sufficiently small neighborhoods $M_1(0;0)$ we have $x^2-2< 0$, посему $x^2(x^2-2) < 0$, откуда следует $x^2(x^2-2)+3 < 3$. Вывод: любая окрестность точки $M_1(0;0)$ содержит точки, в которых $z < 3$, посему точка $M_1(0;0)$ не может быть точкой минимума.

But maybe the point $M_1(0;0)$ is a maximum point? If this is so, then for any point $M$ from some neighborhood of the point $M_1(0;0)$ we get $z(M)< z(M_1) $, т.е. $z(M) < 3$. А вдруг любая окрестность содержит точки, в которых $z(M) >3$? Then there will definitely not be a maximum at the point $M_1$.

Consider points for which $y=x$, i.e. points of the form $(x,x)$. At these points, the $z$ function will take on the following values:

$$ z(x,x)=x^4+x^4-2x^2+4x\cdot x-2\cdot x^2+3=2x^4+3. $$

Since in any neighborhood of the point $M_1(0;0)$ we have $2x^4 > 0$, then $2x^4+3 > 3$. Conclusion: any neighborhood of the point $M_1(0;0)$ contains points where $z > 3$, so the point $M_1(0;0)$ cannot be a maximum point.

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

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

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 exploration and plotting. The extremum point is used when finding the largest and smallest values ​​of the function, since they increase or decrease the function from the interval.

This article reveals the definitions, we formulate a sufficient sign of increase and decrease on the interval and the condition for the existence of an extremum. This applies to solving examples and problems. The section on differentiation of functions should be repeated, because when solving it will be necessary to use finding the derivative.

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) will be feasible. 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 feasible. 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 ascending and descending interval, i.e. (a; b) where x = a, x = b, the points are included in the ascending and descending interval. This does not contradict the definition, which means that it takes place on the interval x.

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

Definition 3

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

The point x 0 is called the minimum point for the function y \u003d f (x) when for all values ​​of x the inequality f (x 0) ≤ f (x) is true. Feature Minimum is the value of the function at the point, and has the notation of the form y m i n .

The 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 the 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 equals the maximum value of the function, and the second figure is more like finding a maximum point at x = b.

Sufficient conditions for increasing and decreasing functions

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

The first sufficient condition for an extremum

Definition 4

Let a function y = f (x) be given, which is differentiable in the ε neighborhood of the point x 0 , and has continuity at the given point x 0 . Hence 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 sign setting conditions:

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

To correctly determine the maximum and minimum points of the 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 the points where the function changes sign.

Consider the algorithm on the example of solving several examples of finding the extrema of the function.

Example 1

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

Decision

The domain of this function is all real numbers except x = 2. First, we 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 \u003d - 1, x \u003d 5, x \u003d 2, that is, each bracket must be equated to zero. Mark on the number line and get:

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

We get that

y "(- 2) \u003d 2 (x + 1) (x - 5) (x - 2) 2 x \u003d - 2 \u003d 2 (- 2 + 1) (- 2 - 5) (- 2 - 2) 2 \u003d 2 7 16 \u003d 7 8\u003e 0, therefore, the interval - ∞; - 1 has a positive derivative. Similarly, we obtain 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 segment will be negative. The third with a minus, the fourth with a plus. To determine continuity, it is necessary to pay attention to the sign of the derivative, if it changes, then this is an extremum point.

We get 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 the 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 +. Hence, x=-1 is the minimum point, and its finding 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 sign of an extremum does not require the function to be differentiable from the point x 0 , and 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 .

Decision.

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 has no 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 perform calculations to find the value of the argument when the derivative becomes 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 points obtained must be marked on the 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) \u003d - 1 2 x 2 - 4 x - 22 3 x \u003d - 6 \u003d - 1 2 - 6 2 - 4 (- 6) - 22 3 \u003d - 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 a straight line has the form

So, we come to the point that it is necessary to resort to the first sign of an extremum. We calculate and get 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 us 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 the function f "(x 0) = 0 is given, then with its f "" (x 0) > 0 we get that x 0 is the 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 .

Decision

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

When x = 1, the derivative becomes equal to zero, which means that the point is a possible extremum. For clarification, it is necessary to find the second derivative and calculate the value at x \u003d 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

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

Graphic image

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

Definition 5

The function y = f (x) has its derivative up to the nth order in the ε neighborhood of the 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)(x0)< 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 .

Decision

The original function is an entire rational one, hence it follows that the domain of definition is all real numbers. The function needs to be differentiated. 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 points of a possible extremum. It is necessary to apply the third sufficient extremum condition. 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 \u003d 5 7 is the maximum point. Applying 3 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, 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

Hence, 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 the above, we conclude that x 3 \u003d 3 is the minimum point of the function.

Graphic image

Answer: x 2 \u003d 5 7 is the maximum point, x 3 \u003d 3 - the minimum point of the given function.

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Introduction

In many fields of science and in practice, one often encounters the problem of finding the extremum of a function. The fact is that many technical, economic, etc. processes are modeled by a function or several functions that depend on variables - factors that affect the state of the phenomenon being modeled. It is required to find the extrema of such functions in order to determine the optimal (rational) state, process control. So in the economy, the problems of minimizing costs or maximizing profits are often solved - the microeconomic task of the firm. In this work, we do not consider modeling issues, but consider only algorithms for finding function extrema in the simplest version, when no restrictions are imposed on variables (unconditional optimization), and the extremum is sought for only one objective function.

EXTREMA OF THE FUNCTION

Consider the graph of a continuous function y=f(x) shown in the figure. Function value at point x 1 will be greater than the values ​​of the function at all neighboring points both to the left and to the right of x one . In this case, the function is said to have at the point x 1 max. At the point x The 3 function obviously also has a maximum. If we consider the point x 2 , then the value of the function in it is less than all neighboring values. In this case, the function is said to have at the point x 2 minimum. Similarly for the point x 4 .

Function y=f(x) at the point x 0 has maximum, if the value of the function at this point is greater than its values ​​at all points of some interval containing the point x 0 , i.e. if there is such a neighborhood of the point x 0 , which is for everyone x≠x 0 , belonging to this neighborhood, we have the inequality f(x)<f(x 0 ) .

Function y=f(x) It has minimum at the point x 0 , if there is such a neighborhood of the point x 0 , what is for everyone x≠x 0 belonging to this neighborhood, we have the inequality f(x)>f(x0.

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

Let us pay attention to the fact that a function defined on a segment can reach its maximum and minimum only at points enclosed within the segment under consideration.

Note that if a function has a maximum at a point, this does not mean that at this point the function has the maximum value in the entire domain of definition. In the figure discussed above, the function at the point x 1 has a maximum, although there are points at which the values ​​of the function are greater than at the point x 1 . In particular, f(x 1) < f(x 4) i.e. the minimum of the function is greater than the maximum. From the definition of the maximum, it only follows that this is the largest value of the function at points sufficiently close to the maximum point.

Theorem 1. (A necessary condition for the existence of an extremum.) If a differentiable function y=f(x) has at the point x=x 0 extremum, then its derivative at this point vanishes.

Proof. Let, for definiteness, at the point x 0 the function has a maximum. Then for sufficiently small increments Δ x we have f(x 0 + Δ x) 0 ) , i.e.

But then

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.

has no derivative at x=0, since it goes to infinity when x=0. But at this point, the function has a maximum. has no derivative at x=0, because at x→0. At this point, the function has neither a maximum nor a minimum. Really, f(x)=0 and at x<0f(x)<0, а при x>0f(x)>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;

at x<x 0 and f "(x)> 0 at x > x 0 , then x 0 is the minimum 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 be x< x 0 . Then c< x 0 and f "(c)> 0. So f "(c)(x-x 0)< 0 and, therefore,

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

Let be x > x 0 . Then c> x 0 and f"(c)< 0. Means f "(c)(x-x 0)< 0. So 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 be 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.