Extrema, largest and smallest values ​​of functions. Label: local maximum

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LOCAL MAXIMUM

LOCAL MAXIMUM

(local maximum) The value of a function that is greater than any adjacent value of its argument or set of arguments, dy/dx= 0 is a necessary condition for achieving a local maximum y=f(x); If this condition is met, a sufficient condition for achieving a local maximum is d2y/dx2 0. Local maximum can also be absolute maximum if no value exists X, at which at more. However, this may not always be the case. Consider the function y = x3–3x.dy/dx = 0 when x2= 1; And d2y/dx2=6x. at has a maximum at x =– 1, but this is only a local, not an absolute maximum, since at can become infinitely large when given a large enough positive value X. See also: figure for the article maximum (maximum).


Economy. Dictionary. - M.: "INFRA-M", Publishing House "Ves Mir". J. Black. General editor: Doctor of Economics Osadchaya I.M.. 2000 .


Economic dictionary. 2000 .

See what "LOCAL MAXIMUM" is in other dictionaries:

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$E \subset \mathbb(R)^(n)$. They say $f$ has local maximum at the point $x_(0) \in E$, if there is a neighborhood $U$ of the point $x_(0)$ such that for all $x \in U$ the inequality $f\left(x\right) \leqslant f is satisfied \left(x_(0)\right)$.

The local maximum is called strict , if the neighborhood $U$ can be chosen so that for all $x \in U$ different from $x_(0)$ there is $f\left(x\right)< f\left(x_{0}\right)$.

Definition
Let $f$ be a real function on the open set $E \subset \mathbb(R)^(n)$. They say $f$ has local minimum at the point $x_(0) \in E$, if there is a neighborhood $U$ of the point $x_(0)$ such that the inequality $f\left(x\right) \geqslant f holds for all $x \in U$ \left(x_(0)\right)$.

A local minimum is called strict if a neighborhood $U$ can be chosen so that for all $x \in U$ different from $x_(0)$ there is $f\left(x\right) > f\left(x_( 0)\right)$.

Local extremum combines the concepts of local minimum and local maximum.

Theorem (necessary condition for the extremum of a differentiable function)
Let $f$ be a real function on the open set $E \subset \mathbb(R)^(n)$. If at the point $x_(0) \in E$ the function $f$ has a local extremum at this point, then $$\text(d)f\left(x_(0)\right)=0.$$ Equal to zero differential is equivalent to the fact that all are equal to zero, i.e. $$\displaystyle\frac(\partial f)(\partial x_(i))\left(x_(0)\right)=0.$$

In the one-dimensional case this is – . Let us denote $\phi \left(t\right) = f \left(x_(0)+th\right)$, where $h$ is an arbitrary vector. The function $\phi$ is defined for values ​​of $t$ that are sufficiently small in absolute value. In addition, with respect to , it is differentiable, and $(\phi)’ \left(t\right) = \text(d)f \left(x_(0)+th\right)h$.
Let $f$ have a local maximum at point x $0$. This means that the function $\phi$ at $t = 0$ has a local maximum and, by Fermat’s theorem, $(\phi)’ \left(0\right)=0$.
So, we got that $df \left(x_(0)\right) = 0$, i.e. function $f$ at point $x_(0)$ equal to zero on any vector $h$.

Definition
Points at which the differential is zero, i.e. those in which all partial derivatives are equal to zero are called stationary. Critical points functions $f$ are those points at which $f$ is not differentiable or is equal to zero. If the point is stationary, then it does not follow from this that the function has an extremum at this point.

Example 1.
Let $f \left(x,y\right)=x^(3)+y^(3)$. Then $\displaystyle\frac(\partial f)(\partial x) = 3 \cdot x^(2)$,$\displaystyle\frac(\partial f)(\partial y) = 3 \cdot y^(2 )$, so $\left(0,0\right)$ is a stationary point, but the function has no extremum at this point. Indeed, $f \left(0,0\right) = 0$, but it is easy to see that in any neighborhood of the point $\left(0,0\right)$ the function takes both positive and negative values.

Example 2.
The function $f \left(x,y\right) = x^(2) − y^(2)$ has a stationary point at its origin, but it is clear that there is no extremum at this point.

Theorem (sufficient condition for extremum).
Let the function $f$ be twice continuously differentiable on the open set $E \subset \mathbb(R)^(n)$. Let $x_(0) \in E$ be a stationary point and $$\displaystyle Q_(x_(0)) \left(h\right) \equiv \sum_(i=1)^n \sum_(j=1) ^n \frac(\partial^(2) f)(\partial x_(i) \partial x_(j)) \left(x_(0)\right)h^(i)h^(j).$$ Then

  1. if $Q_(x_(0))$ – , then the function $f$ at the point $x_(0)$ has a local extremum, namely, a minimum if the form is positive definite, and a maximum if the form is negative definite;
  2. If quadratic form$Q_(x_(0))$ is undefined, then the function $f$ at the point $x_(0)$ has no extremum.

Let's use the expansion according to Taylor's formula (12.7 p. 292). Considering that the first order partial derivatives at the point $x_(0)$ are equal to zero, we obtain $$\displaystyle f \left(x_(0)+h\right)−f \left(x_(0)\right) = \ frac(1)(2) \sum_(i=1)^n \sum_(j=1)^n \frac(\partial^(2) f)(\partial x_(i) \partial x_(j)) \left(x_(0)+\theta h\right)h^(i)h^(j),$$ where $0<\theta<1$. Обозначим $\displaystyle a_{ij}=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right)$. В силу теоремы Шварца (12.6 стр. 289-290) , $a_{ij}=a_{ji}$. Обозначим $$\displaystyle \alpha_{ij} \left(h\right)=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}+\theta h\right)−\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right).$$ По предположению, все непрерывны и поэтому $$\lim_{h \rightarrow 0} \alpha_{ij} \left(h\right)=0. \left(1\right)$$ Получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left.$$ Обозначим $$\displaystyle \epsilon \left(h\right)=\frac{1}{|h|^{2}}\sum_{i=1}^n \sum_{j=1}^n \alpha_{ij} \left(h\right)h_{i}h_{j}.$$ Тогда $$|\epsilon \left(h\right)| \leq \sum_{i=1}^n \sum_{j=1}^n |\alpha_{ij} \left(h\right)|$$ и, в силу соотношения $\left(1\right)$, имеем $\epsilon \left(h\right) \rightarrow 0$ при $h \rightarrow 0$. Окончательно получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left. \left(2\right)$$ Предположим, что $Q_{x_{0}}$ – положительноопределенная форма. Согласно лемме о положительноопределённой квадратичной форме (12.8.1 стр. 295, Лемма 1) , существует такое положительное число $\lambda$, что $Q_{x_{0}} \left(h\right) \geqslant \lambda|h|^{2}$ при любом $h$. Поэтому $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right) \geq \frac{1}{2}|h|^{2} \left(λ+\epsilon \left(h\right)\right).$$ Так как $\lambda>0$, and $\epsilon \left(h\right) \rightarrow 0$ for $h \rightarrow 0$, then the right-hand side will be positive for any vector $h$ of sufficiently small length.
So, we have come to the conclusion that in a certain neighborhood of the point $x_(0)$ the inequality $f \left(x\right) >f \left(x_(0)\right)$ holds if only $x \neq x_ (0)$ (we put $x=x_(0)+h$\right). This means that at the point $x_(0)$ the function has a strict local minimum, and thus the first part of our theorem is proved.
Let us now assume that $Q_(x_(0))$ is an indefinite form. Then there are vectors $h_(1)$, $h_(2)$ such that $Q_(x_(0)) \left(h_(1)\right)=\lambda_(1)>0$, $Q_ (x_(0)) \left(h_(2)\right)= \lambda_(2)<0$. В соотношении $\left(2\right)$ $h=th_{1}$ $t>$0. Then we get $$f \left(x_(0)+th_(1)\right)−f \left(x_(0)\right) = \frac(1)(2) \left[ t^(2) \ lambda_(1) + t^(2) |h_(1)|^(2) \epsilon \left(th_(1)\right) \right] = \frac(1)(2) t^(2) \ left[ \lambda_(1) + |h_(1)|^(2) \epsilon \left(th_(1)\right) \right].$$ For sufficiently small $t>0$, the right-hand side is positive. This means that in any neighborhood of the point $x_(0)$ the function $f$ takes values ​​$f \left(x\right)$ greater than $f \left(x_(0)\right)$.
Similarly, we find that in any neighborhood of the point $x_(0)$ the function $f$ takes values ​​less than $f \left(x_(0)\right)$. This, together with the previous one, means that at the point $x_(0)$ the function $f$ does not have an extremum.

Let's consider special case of this theorem for a function $f \left(x,y\right)$ of two variables defined in a certain neighborhood of the point $\left(x_(0),y_(0)\right)$ and having continuous partial derivatives of the first in this neighborhood and second orders. Assume that $\left(x_(0),y_(0)\right)$ is a stationary point and denote $$\displaystyle a_(11)= \frac(\partial^(2) f)(\partial x ^(2)) \left(x_(0) ,y_(0)\right), a_(12)=\frac(\partial^(2) f)(\partial x \partial y) \left(x_( 0), y_(0)\right), a_(22)=\frac(\partial^(2) f)(\partial y^(2)) \left(x_(0), y_(0)\right ).$$ Then the previous theorem takes the following form.

Theorem
Let $\Delta=a_(11) \cdot a_(22) − a_(12)^2$. Then:

  1. if $\Delta>0$, then the function $f$ has a local extremum at the point $\left(x_(0),y_(0)\right)$, namely, a minimum if $a_(11)>0$ , and maximum if $a_(11)<0$;
  2. if $\Delta<0$, то экстремума в точке $\left(x_{0},y_{0}\right)$ нет. Как и в одномерном случае, при $\Delta=0$ экстремум может быть, а может и не быть.

Examples of problem solving

Algorithm for finding the extremum of a function of many variables:

  1. Finding stationary points;
  2. Find the 2nd order differential at all stationary points
  3. Using the sufficient condition for the extremum of a function of many variables, we consider the 2nd order differential at each stationary point
  1. Investigate the function for extremum $f \left(x,y\right) = x^(3) + 8 \cdot y^(3) + 18 \cdot x — 30 \cdot y$.
    Solution

    Let's find the 1st order partial derivatives: $$\displaystyle \frac(\partial f)(\partial x)=3 \cdot x^(2) - 6 \cdot y;$$ $$\displaystyle \frac(\partial f)(\partial y)=24 \cdot y^(2) - 6 \cdot x.$$ Let's compose and solve the system: $$\displaystyle \begin(cases)\frac(\partial f)(\partial x) = 0\\\frac(\partial f)(\partial y)= 0\end(cases) \Rightarrow \begin(cases)3 \cdot x^(2) - 6 \cdot y= 0\\24 \cdot y^(2) — 6 \cdot x = 0\end(cases) \Rightarrow \begin(cases)x^(2) — 2 \cdot y= 0\\4 \cdot y^(2) — x = 0 \end(cases)$$ From the 2nd equation we express $x=4 \cdot y^(2)$ - substitute it into the 1st equation: $$\displaystyle \left(4 \cdot y^(2)\right )^(2)-2 \cdot y=0$$ $$16 \cdot y^(4) — 2 \cdot y = 0$$ $$8 \cdot y^(4) — y = 0$$ $$y \left(8 \cdot y^(3) -1\right)=0$$ As a result, 2 stationary points are obtained:
    1) $y=0 \Rightarrow x = 0, M_(1) = \left(0, 0\right)$;
    2) $\displaystyle 8 \cdot y^(3) -1=0 \Rightarrow y^(3)=\frac(1)(8) \Rightarrow y = \frac(1)(2) \Rightarrow x=1 , M_(2) = \left(\frac(1)(2), 1\right)$
    Let's check whether the sufficient condition for an extremum is satisfied:
    $$\displaystyle \frac(\partial^(2) f)(\partial x^(2))=6 \cdot x; \frac(\partial^(2) f)(\partial x \partial y)=-6; \frac(\partial^(2) f)(\partial y^(2))=48 \cdot y$$
    1) For the point $M_(1)= \left(0,0\right)$:
    $$\displaystyle A_(1)=\frac(\partial^(2) f)(\partial x^(2)) \left(0,0\right)=0; B_(1)=\frac(\partial^(2) f)(\partial x \partial y) \left(0,0\right)=-6; C_(1)=\frac(\partial^(2) f)(\partial y^(2)) \left(0,0\right)=0;$$
    $A_(1) \cdot B_(1) — C_(1)^(2) = -36<0$ , значит, в точке $M_{1}$ нет экстремума.
    2) For point $M_(2)$:
    $$\displaystyle A_(2)=\frac(\partial^(2) f)(\partial x^(2)) \left(1,\frac(1)(2)\right)=6; B_(2)=\frac(\partial^(2) f)(\partial x \partial y) \left(1,\frac(1)(2)\right)=-6; C_(2)=\frac(\partial^(2) f)(\partial y^(2)) \left(1,\frac(1)(2)\right)=24;$$
    $A_(2) \cdot B_(2) — C_(2)^(2) = 108>0$, which means that at point $M_(2)$ there is an extremum, and since $A_(2)>0$, then this is the minimum.
    Answer: The point $\displaystyle M_(2)\left(1,\frac(1)(2)\right)$ is the minimum point of the function $f$.

  2. Investigate the function for the extremum $f=y^(2) + 2 \cdot x \cdot y - 4 \cdot x - 2 \cdot y - 3$.
    Solution

    Let's find stationary points: $$\displaystyle \frac(\partial f)(\partial x)=2 \cdot y - 4;$$ $$\displaystyle \frac(\partial f)(\partial y)=2 \cdot y + 2 \cdot x — 2.$$
    Let's compose and solve the system: $$\displaystyle \begin(cases)\frac(\partial f)(\partial x)= 0\\\frac(\partial f)(\partial y)= 0\end(cases) \ Rightarrow \begin(cases)2 \cdot y - 4= 0\\2 \cdot y + 2 \cdot x - 2 = 0\end(cases) \Rightarrow \begin(cases) y = 2\\y + x = 1\end(cases) \Rightarrow x = -1$$
    $M_(0) \left(-1, 2\right)$ is a stationary point.
    Let's check whether the sufficient condition for the extremum is met: $$\displaystyle A=\frac(\partial^(2) f)(\partial x^(2)) \left(-1,2\right)=0; B=\frac(\partial^(2) f)(\partial x \partial y) \left(-1,2\right)=2; C=\frac(\partial^(2) f)(\partial y^(2)) \left(-1,2\right)=2;$$
    $A \cdot B — C^(2) = -4<0$ , значит, в точке $M_{0}$ нет экстремума.
    Answer: there are no extremes.

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    Task 1 of 4

    1 .
    Number of points: 1

    Investigate the function $f$ for extrema: $f=e^(x+y)(x^(2)-2 \cdot y^(2))$

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  1. Task 2 of 4

    2 .
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    Does the function $f = 4 + \sqrt((x^(2)+y^(2))^(2))$ have an extremum

The extremum point of a function is the point in the domain of definition of the function at which the value of the function takes on a minimum or maximum value. The values ​​of the function at these points are called extrema (minimum and maximum) of the function.

Definition. Dot x1 function domain f(x) is called maximum point of the function , if the value of the function at this point is greater than the values ​​of the function at points sufficiently close to it, located to the right and left of it (that is, the inequality holds f(x0 ) > f(x 0 + Δ x) x1 maximum.

Definition. Dot x2 function domain f(x) is called minimum point of the function, if the value of the function at this point is less than the values ​​of the function at points sufficiently close to it, located to the right and left of it (that is, the inequality holds f(x0 ) < f(x 0 + Δ x) ). In this case we say that the function has at the point x2 minimum.

Let's say point x1 - maximum point of the function f(x) . Then in the interval up to x1 function increases, therefore the derivative of the function is greater than zero ( f "(x) > 0 ), and in the interval after x1 the function decreases, therefore, derivative of a function less than zero (f "(x) < 0 ). Тогда в точке x1

Let us also assume that the point x2 - minimum point of the function f(x) . Then in the interval up to x2 the function is decreasing, and the derivative of the function is less than zero ( f "(x) < 0 ), а в интервале после x2 the function is increasing, and the derivative of the function is greater than zero ( f "(x) > 0 ). In this case also at the point x2 the derivative of the function is zero or does not exist.

Fermat's theorem (a necessary sign of the existence of an extremum of a function). If the point x0 - extremum point of the function f(x) then at this point the derivative of the function is equal to zero ( f "(x) = 0 ) or does not exist.

Definition. The points at which the derivative of a function is zero or does not exist are called critical points .

Example 1. Let's consider the function.

At the point x= 0 the derivative of the function is zero, therefore the point x= 0 is the critical point. However, as can be seen on the graph of the function, it increases throughout the entire domain of definition, so the point x= 0 is not the extremum point of this function.

Thus, the conditions that the derivative of a function at a point is equal to zero or does not exist are necessary conditions for an extremum, but not sufficient, since other examples of functions can be given for which these conditions are met, but the function does not have an extremum at the corresponding point. That's why there must be sufficient evidence, allowing one to judge whether there is an extremum at a particular critical point and what kind of extremum it is - maximum or minimum.

Theorem (the first sufficient sign of the existence of an extremum of a function). Critical point x0 f(x) if, when passing through this point, the derivative of the function changes sign, and if the sign changes from “plus” to “minus”, then it is a maximum point, and if from “minus” to “plus”, then it is a minimum point.

If near the point x0 , to the left and to the right of it, the derivative retains its sign, this means that the function either only decreases or only increases in a certain neighborhood of the point x0 . In this case, at the point x0 there is no extreme.

So, to determine the extremum points of the function, you need to do the following :

  1. Find the derivative of the function.
  2. Equate the derivative to zero and determine the critical points.
  3. Mentally or on paper, mark the critical points on the number line and determine the signs of the derivative of the function in the resulting intervals. If the sign of the derivative changes from “plus” to “minus”, then the critical point is the maximum point, and if from “minus” to “plus”, then the minimum point.
  4. Calculate the value of the function at the extremum points.

Example 2. Find the extrema of the function .

Solution. Let's find the derivative of the function:

Let's equate the derivative to zero to find the critical points:

.

Since for any values ​​of “x” the denominator is not equal to zero, we equate the numerator to zero:

Got one critical point x= 3 . Let us determine the sign of the derivative in the intervals delimited by this point:

in the range from minus infinity to 3 - a minus sign, that is, the function decreases,

in the interval from 3 to plus infinity there is a plus sign, that is, the function increases.

That is, period x= 3 is the minimum point.

Let's find the value of the function at the minimum point:

Thus, the extremum point of the function is found: (3; 0), and it is the minimum point.

Theorem (the second sufficient sign of the existence of an extremum of a function). Critical point x0 is the extremum point of the function f(x) if the second derivative of the function at this point is not equal to zero ( f ""(x) ≠ 0 ), and if the second derivative is greater than zero ( f ""(x) > 0 ), then the maximum point, and if the second derivative is less than zero ( f ""(x) < 0 ), то точкой минимума.

Note 1. If at the point x0 If both the first and second derivatives vanish, then at this point it is impossible to judge the presence of an extremum based on the second sufficient criterion. In this case, you need to use the first sufficient criterion for the extremum of a function.

Remark 2. The second sufficient criterion for the extremum of a function is not applicable even when the first derivative does not exist at a stationary point (then the second derivative does not exist either). In this case, you also need to use the first sufficient sign of an extremum of a function.

Local nature of the extrema of the function

From the above definitions it follows that the extremum of a function is local in nature - it is the largest and smallest value of the function compared to nearby values.

Let's say you're looking at your earnings over a period of one year. If in May you earned 45,000 rubles, and in April 42,000 rubles and in June 39,000 rubles, then May earnings are the maximum of the earnings function compared to nearby values. But in October you earned 71,000 rubles, in September 75,000 rubles, and in November 74,000 rubles, so October earnings are the minimum of the earnings function compared to nearby values. And you can easily see that the maximum among the values ​​of April-May-June is less than the minimum of September-October-November.

Generally speaking, on an interval a function can have several extrema, and it may turn out that some minimum of the function is greater than any maximum. So, for the function shown in the figure above, .

That is, one should not think that the maximum and minimum of a function are, respectively, its largest and smallest values ​​on the entire segment under consideration. At the maximum point, the function has the greatest value only in comparison with those values ​​​​that it has at all points sufficiently close to the maximum point, and at the minimum point it has the smallest value only in comparison with those values ​​​​that it has at all points sufficiently close to the minimum point.

Therefore, we can clarify the above concept of extremum points of a function and call minimum points local minimum points, and maximum points local maximum points.

We look for the extrema of the function together

Example 3.

Solution: The function is defined and continuous on the entire number line. Its derivative also exists on the entire number line. Therefore, in this case, the critical points are only those at which, i.e. , from where and . Critical points and divide the entire domain of definition of the function into three intervals of monotonicity: . Let us select one control point in each of them and find the sign of the derivative at this point.

For the interval, the control point can be: find. Taking a point in the interval, we get, and taking a point in the interval, we have. So, in the intervals and , and in the interval . According to the first sufficient criterion for an extremum, there is no extremum at the point (since the derivative retains its sign in the interval), and at the point the function has a minimum (since the derivative changes sign from minus to plus when passing through this point). Let's find the corresponding values ​​of the function: , a . In the interval the function decreases, since in this interval , and in the interval it increases, since in this interval .

To clarify the construction of the graph, we find the points of intersection of it with the coordinate axes. When we obtain an equation whose roots are and , i.e., two points (0; 0) and (4; 0) of the graph of the function are found. Using all the information received, we build a graph (see the beginning of the example).

Example 4. Find the extrema of the function and build its graph.

The domain of definition of a function is the entire number line, except for the point, i.e. .

To shorten the study, you can use the fact that this function is even, since . Therefore, its graph is symmetrical about the axis Oy and the study can only be performed for the interval.

Finding the derivative and critical points of the function:

1) ;

2) ,

but the function suffers a discontinuity at this point, so it cannot be an extremum point.

Thus, the given function has two critical points: and . Taking into account the parity of the function, we will check only the point using the second sufficient criterion for an extremum. To do this, we find the second derivative and determine its sign at: we get . Since and , it is the minimum point of the function, and .

To get a more complete picture of the graph of a function, let’s find out its behavior at the boundaries of the domain of definition:

(here the symbol indicates the desire x to zero from the right, and x remains positive; similarly means aspiration x to zero from the left, and x remains negative). Thus, if , then . Next, we find

,

those. if , then .

The graph of a function has no intersection points with the axes. The picture is at the beginning of the example.

We continue to search for extrema of the function together

Example 8. Find the extrema of the function.

Solution. Let's find the domain of definition of the function. Since the inequality must be satisfied, we obtain from .

Let's find the first derivative of the function:

Let's find the critical points of the function.

The function is said to have at the internal point
region D local maximum(minimum), if there is such a neighborhood of the point
, for each point
which holds the inequality

If a function has at a point
local maximum or local minimum, then we say that it has at this point local extremum(or just an extreme).

Theorem (necessary condition for the existence of an extremum). If the differentiable function reaches an extremum at the point
, then each first-order partial derivative of the function at this point it becomes zero.

The points at which all first-order partial derivatives vanish are called stationary points of the function
. The coordinates of these points can be found by solving the system of equations

.

The necessary condition for the existence of an extremum in the case of a differentiable function can be briefly formulated as follows:

There are cases when at individual points some partial derivatives have infinite values ​​or do not exist (while the rest are equal to zero). Such points are called critical points of the function. These points should also be considered as “suspicious” for an extremum, just like stationary ones.

In the case of a function of two variables, the necessary condition for the extremum, namely the equality to zero of the partial derivatives (differential) at the extremum point, has a geometric interpretation: tangent plane to the surface
at the extremum point must be parallel to the plane
.

20. Sufficient conditions for the existence of an extremum

Execution at some point necessary condition the existence of an extremum does not at all guarantee the presence of an extremum there. As an example, we can take the everywhere differentiable function
. Both of its partial derivatives and the function itself vanish at the point
. However, in any neighborhood of this point there are both positive (large
), and negative (smaller
) values ​​of this function. Therefore, at this point, by definition, no extremum is observed. Therefore it is necessary to know sufficient conditions, at which the point suspected of being an extremum is the extremum point of the function under study.

Let's consider the case of a function of two variables. Let's assume that the function
defined, continuous and has continuous partial derivatives up to the second order inclusive in the neighborhood of some point
, which is the stationary point of the function
, that is, satisfies the conditions

,
.

Let us introduce the following notation:

Theorem (sufficient conditions for the existence of an extremum). Let the function
satisfies the above conditions, namely: it is differentiable in some neighborhood of a stationary point
and is twice differentiable at the point itself
. Then if


In case
then the function
at the point
reaches

local maximum at
And

local minimum at
.

In general, for the function
sufficient condition for existence at the point
localminimum(maximum) is positive(negative) certainty of the second differential.

In other words, the following statement is true.

Theorem . If at the point
for function

for any not equal to zero at the same time
, then at this point the function has minimum(similar to maximum, If
).

Example 18.Find local extremum points of a function

Solution. Let's find the partial derivatives of the function and equate them to zero:

Solving this system, we find two possible extremum points:

Let's find the second order partial derivatives for this function:

At the first stationary point, therefore, and
Therefore, additional research is required at this point. Function value
at this point is zero:
Next,

at

A

at

Therefore, in any neighborhood of the point
function
takes values ​​as large
, and smaller
, and, therefore, at the point
function
, by definition, has no local extremum.

At the second stationary point



therefore, therefore, since
then at the point
the function has a local maximum.

The function increments to the argument increment, which tends to zero. To find it, use the table of derivatives. For example, the derivative of the function y = x3 will be equal to y’ = x2.

Equate this derivative to zero (in this case x2=0).

Find the value of the given variable. These will be the values ​​at which the given derivative will be equal to 0. To do this, substitute arbitrary numbers in the expression instead of x, at which the entire expression will become zero. For example:

2-2x2= 0
(1-x)(1+x) = 0
x1= 1, x2 = -1

Plot the obtained values ​​on the coordinate line and calculate the sign of the derivative for each of the obtained values. Points are marked on the coordinate line, which are taken as the origin. To calculate the value in the intervals, substitute arbitrary values ​​that match the criteria. For example, for the previous function before the interval -1, you can select the value -2. For values ​​from -1 to 1, you can select 0, and for values ​​greater than 1, select 2. Substitute these numbers into the derivative and find out the sign of the derivative. In this case, the derivative with x = -2 will be equal to -0.24, i.e. negative and there will be a minus sign on this interval. If x=0, then the value will be equal to 2, and a sign is placed on this interval. If x=1, then the derivative will also be equal to -0.24 and a minus is put.

If, when passing through a point on the coordinate line, the derivative changes its sign from minus to plus, then this is a minimum point, and if from plus to minus, then this is a maximum point.

Video on the topic

Useful advice

To find the derivative, there are online services that calculate the required values ​​and display the result. On such sites you can find derivatives up to 5th order.

Sources:

  • One of the services for calculating derivatives
  • maximum point of the function

The maximum points of a function, along with the minimum points, are called extremum points. At these points the function changes its behavior. Extrema are determined on limited numerical intervals and are always local.

Instructions

Finding process local extremes is called a function and is performed by analyzing the first and second derivatives of the function. Before starting the study, make sure that the specified range of argument values ​​belongs to acceptable values. For example, for the function F=1/x the argument x=0 is not valid. Or for the function Y=tg(x) the argument cannot have the value x=90°.

Make sure that the function Y is differentiable over the entire given interval. Find the first derivative of Y." Obviously, before reaching the point of local maximum, the function increases, and when passing through the maximum, the function becomes decreasing. The first derivative in its physical meaning characterizes the rate of change of a function. While the function is increasing, the rate of this process is positive. When passing through a local maximum, the function begins to decrease, and the rate of change of the function becomes negative. The transition of the rate of change of the function through zero occurs at the point of local maximum.

For example, the function Y=-x²+x+1 on the segment from -1 to 1 has a continuous derivative Y"=-2x+1. At x=1/2 the derivative is zero, and when passing through this point the derivative changes sign from " +" to "-". The second derivative of the function Y"=-2. Plot a point-by-point graph of the function Y=-x²+x+1 and check whether the point with the abscissa x=1/2 is a local maximum on a given segment of the number axis.

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