How to determine the minimum and maximum points of a function. What are extrema of a function: critical points of maximum and minimum

From this article the reader will learn about what an extremum of a functional value is, as well as about the features of its use in practical activities. Studying such a concept is extremely important for understanding the foundations of higher mathematics. This topic is fundamental for a deeper study of the course.

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What is an extremum?

In the school course, many definitions of the concept “extremum” are given. This article is intended to give the deepest and clearest understanding of the term for those ignorant of the issue. So, the term is understood to what extent the functional interval acquires a minimum or maximum value on a particular set.

An extremum is both the minimum value of a function and the maximum at the same time. There is a minimum point and a maximum point, that is, the extreme values ​​of the argument on the graph. The main sciences that use this concept are:

• statistics;
• machine control;
• econometrics.

Extremum points play an important role in determining the sequence of a given function. The coordinate system in the graph at its best shows the change in extreme position depending on the change in functionality.

Extrema of the derivative function

There is also such a phenomenon as “derivative”. It is necessary to determine the extremum point. It is important not to confuse minimum or maximum points with the highest and lowest values. These are different concepts, although they may seem similar.

The value of the function is the main factor in determining how to find the maximum point. The derivative is not formed from values, but exclusively from its extreme position in one or another order.

The derivative itself is determined based on these extremum points, and not on the largest or smallest value. In Russian schools, the line between these two concepts is not clearly drawn, which affects the understanding of this topic in general.

Let us now consider such a concept as “acute extremum”. Today, there is an acute minimum value and an acute maximum value. The definition is given in accordance with the Russian classification of critical points of a function. The concept of an extremum point is the basis for finding critical points on a graph.

To define such a concept, they resort to using Fermat's theorem. It is the most important in the study of extreme points and gives a clear idea of ​​their existence in one form or another. To ensure extremeness, it is important to create certain conditions for a decrease or increase on the graph.

To accurately answer the question “how to find the maximum point”, you must follow these guidelines:

1. Finding the exact domain of definition on the graph.
2. Search for the derivative of a function and the extremum point.
3. Solve standard inequalities for the domain where the argument is found.
4. Be able to prove in which functions a point on a graph is defined and continuous.

Attention! Searching for the critical point of a function is possible only if there is a derivative of at least second order, which is ensured by a high proportion of the presence of an extremum point.

Necessary condition for the extremum of a function

In order for an extremum to exist, it is important that there are both minimum and maximum points. If this rule is only partially observed, then the condition for the existence of an extremum is violated.

Each function in any position must be differentiated in order to identify its new meanings. It is important to understand that the case of a point going to zero is not the main principle for finding a differentiable point.

An acute extremum, as well as a minimum of a function, is an extremely important aspect of solving a mathematical problem using extreme values. In order to better understand this component, it is important to refer to the tabular values ​​for specifying the functionality.

The function and the study of its features occupies one of the key chapters in modern mathematics. The main component of any function is graphs depicting not only its properties, but also the parameters of the derivative of this function. Let's understand this difficult topic. So what is the best way to find the maximum and minimum points of a function?

Function: definition

Any variable that in some way depends on the values ​​of another quantity can be called a function. For example, the function f(x 2) is quadratic and determines the values ​​for the entire set x. Let's say that x = 9, then the value of our function will be equal to 9 2 = 81.

Functions come in many different types: logical, vector, logarithmic, trigonometric, numeric and others. They were studied by such outstanding minds as Lacroix, Lagrange, Leibniz and Bernoulli. Their works serve as a mainstay in modern ways of studying functions. Before finding the minimum points, it is very important to understand the very meaning of the function and its derivative.

Derivative and its role

All functions depend on their variables, which means that they can change their value at any time. On the graph, this will be depicted as a curve that either falls or rises along the ordinate axis (this is the whole set of “y” numbers along the vertical graph). So, determining the maximum and minimum points of a function is precisely related to these “oscillations”. Let us explain what this relationship is.

The derivative of any function is graphed in order to study its basic characteristics and calculate how quickly the function changes (i.e. changes its value depending on the variable "x"). At the moment when the function increases, the graph of its derivative will also increase, but at any second the function can begin to decrease, and then the graph of the derivative will decrease. Those points at which the derivative changes from a minus sign to a plus sign are called minimum points. In order to know how to find minimum points, you should better understand

How to calculate derivative?

The definition and functions imply several concepts from In general, the very definition of a derivative can be expressed as follows: this is the quantity that shows the rate of change of the function.

The mathematical way of determining it seems complicated for many students, but in reality everything is much simpler. You just need to follow the standard plan for finding the derivative of any function. Below we describe how you can find the minimum point of a function without applying the rules of differentiation and without memorizing the table of derivatives.

1. You can calculate the derivative of a function using a graph. To do this, you need to depict the function itself, then take one point on it (point A in the figure). Draw a line vertically down to the abscissa axis (point x 0), and at point A draw a tangent to the graph of the function. The x-axis and the tangent form a certain angle a. To calculate the value of how quickly a function increases, you need to calculate the tangent of this angle a.
2. It turns out that the tangent of the angle between the tangent and the direction of the x-axis is the derivative of the function in a small area with point A. This method is considered a geometric method for determining the derivative.

Methods for studying function

In the school mathematics curriculum, it is possible to find the minimum point of a function in two ways. We have already discussed the first method using a graph, but how can we determine the numerical value of the derivative? To do this, you will need to learn several formulas that describe the properties of the derivative and help convert variables like “x” into numbers. The following method is universal, so it can be applied to almost all types of functions (both geometric and logarithmic).

1. It is necessary to equate the function to the derivative function, and then simplify the expression using the rules of differentiation.
2. In some cases, when given a function in which the variable “x” is in the divisor, it is necessary to determine the range of acceptable values, excluding the point “0” from it (for the simple reason that in mathematics one should never divide by zero).
3. After this, you should transform the original form of the function into a simple equation, equating the entire expression to zero. For example, if the function looked like this: f(x) = 2x 3 +38x, then according to the rules of differentiation its derivative is equal to f"(x) = 3x 2 +1. Then we transform this expression into an equation of the following form: 3x 2 +1 = 0 .
4. After solving the equation and finding the “x” points, you should plot them on the x-axis and determine whether the derivative in these sections between the marked points is positive or negative. After the designation, it will become clear at what point the function begins to decrease, that is, changes sign from minus to the opposite. It is in this way that you can find both the minimum and maximum points.

Rules of differentiation

The most basic component in studying a function and its derivative is knowledge of the rules of differentiation. Only with their help can you transform cumbersome expressions and large complex functions. Let's get acquainted with them, there are quite a lot of them, but they are all very simple due to the natural properties of both power and logarithmic functions.

1. The derivative of any constant is equal to zero (f(x) = 0). That is, the derivative f(x) = x 5 + x - 160 will take the following form: f" (x) = 5x 4 +1.
2. Derivative of the sum of two terms: (f+w)" = f"w + fw".
3. Derivative of a logarithmic function: (log a d)" = d/ln a*d. This formula applies to all types of logarithms.
4. Derivative of the power: (x n)"= n*x n-1. For example, (9x 2)" = 9*2x = 18x.
5. The derivative of the sinusoidal function: (sin a)" = cos a. If the sin of angle a is 0.5, then its derivative is √3/2.

Extremum points

We have already discussed how to find minimum points, but there is a concept and functions. If the minimum denotes those points at which the function changes from a minus sign to a plus, then the maximum points are those points on the x-axis at which the derivative of the function changes from plus to the opposite - minus.

You can find maximum points using the method described above, but you should take into account that they indicate those areas where the function begins to decrease, that is, the derivative will be less than zero.

In mathematics, it is customary to generalize both concepts, replacing them with the phrase “points of extrema.” When a task asks you to identify these points, it means that you need to calculate the derivative of a given function and find the minimum and maximum points.

Hello, dear friends! We continue to consider tasks related to the study of functions. I recommend it, which is necessary for solving problems of finding the maximum (minimum) value of a function and finding the maximum (minimum) points of a function.

Problems with logarithms to find the largest (smallest) value of the function we. In this article, we will consider three problems in which the question is of finding the maximum (minimum) points of functions, and the given function contains a natural logarithm.

Theoretical point:

By definition of a logarithm, the expression under the logarithm sign must be greater than zero. *This must be taken into account not only in these problems, but also when solving equations and inequalities containing a logarithm.

Algorithm for finding maximum (minimum) points of a function:

1. Calculate the derivative of the function.

2. We equate it to zero and solve the equation.

3. We mark the resulting roots on the number line.*We also mark the points at which the derivative does not exist. Let us obtain the intervals over which the function increases or decreases.

4. Determine the signs of the derivative on these intervals (substituting arbitrary values ​​from them into the derivative).

5. We draw a conclusion.

Find the maximum point of the function y = ln (x–11)–5x+2

Let us immediately write down that x–11>0 (by the definition of a logarithm), that is, x > 11.

We will consider the function on the interval (11;∞).

Let's find the zeros of the derivative:

The point x = 11 is not included in the domain of definition of the function and the derivative does not exist at it. We mark two points on the number axis: 11 and 11.2. Let's determine the signs of the derivative of the function by substituting arbitrary values ​​from the intervals (11;11,2) and (11,2;+∞) into the found derivative, and depict the behavior of the function in the figure:

Thus, at the point x = 11.2, the derivative of the function changes sign from positive to negative, which means this is the desired maximum point.

Answer: 11.2

Decide for yourself:

Find the maximum point of the function y=ln (x+5)–2x+9.

Find the minimum point of the function y=4x– ln (x+5)+8

Let us immediately write down that x+5>0 (by the property of the logarithm), that is, x>–5.

We will consider the function on the interval (– 5;+∞).

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

Point x = –5 is not included in the domain of definition of the function and the derivative does not exist in it. Mark two points on the number axis–5 and –4.75. Let’s determine the signs of the derivative of the function by substituting arbitrary values ​​from the intervals (–5;–4.75) and (–4.75;+∞) into the found derivative, and depict the behavior of the function in the figure:

Thus, at the point x = –4.75, the derivative of the function changes sign from negative to positive, which means this is the desired minimum point.

Answer: – 4.75

Decide for yourself:

Find the minimum point of the function y=2x–ln (x+3)+7.

Find the maximum point of the function y = x 2 –34x+140lnx–10

According to the property of a logarithm, the expression under its sign is greater than zero, that is, x > 0.

We will consider the function on the interval (0; +∞).

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

Solving the quadratic equation, we get: D = 9 x 1 = 10 x 2 = 7.

Point x = 0 is not included in the domain of definition of the function and the derivative does not exist in it. We mark three points on the number axis 0, 7 and 10.

The ox axis is divided into intervals: (0;7), (7;10), (10; +∞).

Let's determine the signs of the derivative of the function by substituting arbitrary values ​​from the obtained intervals into the found derivative, and depict the behavior of the function in the figure:

That's all. I wish you success!

Sincerely, Alexander Krutitskikh

P.S: I would be grateful if you tell me about the site on social networks.

The algorithm for finding these points has already been discussed several times, but I’ll repeat it briefly:

1. Find the derivative of the function.

2. Find the zeros of the derivative (equate the derivative to zero and solve the equation).

3. Next, we build a number line, mark the found points on it and determine the signs of the derivative on the resulting intervals. *This is done by substituting arbitrary values ​​from the intervals into the derivative.

If you are not at all familiar with the properties of derivatives for studying functions, then be sure to study the article« ». Also repeat the table of derivatives and the rules of differentiation (available in the same article). Let's consider the tasks:

77431. Find the maximum point of the function y = x 3 –5x 2 +7x–5.

Let's find the derivative of the function:

Let's find the zeros of the derivative:

3x 2 – 10x + 7 = 0

y(0)" = 3∙0 2 – 10∙0 + 7 = 7 > 0

y(2)" = 3∙2 2 – 10∙2 + 7 = – 1< 0

y(3)" = 3∙3 2 – 10∙3 + 7 = 4 > 0

At the point x = 1, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

Answer: 1

77432. Find the minimum point of the function y = x 3 +5x 2 +7x–5.

Let's find the derivative of the function:

Let's find the zeros of the derivative:

3x 2 + 10x + 7 = 0

Solving the quadratic equation we get:

We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:

y(–3 ) " = 3∙(–3) 2 + 10∙(–3) + 7 = 4 > 0

y(–2 ) "= 3∙(–2) 2 + 10∙(–2) + 7 = –1 < 0

y(0) "= 3∙0 2 – 10∙0 + 7 = 7 > 0

At the point x = –1, the derivative changes its sign from negative to positive, which means this is the desired minimum point.

Answer: –1

77435. Find the maximum point of the function y = 7 + 12x – x 3

Let's find the derivative of the function:

Let's find the zeros of the derivative:

12 – 3x 2 = 0

x 2 = 4

Solving the equation we get:

*These are points of possible maximum (minimum) of the function.

We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:

y(–3 ) "= 12 – 3∙(–3) 2 = –15 < 0

y(0) "= 12 – 3∙0 2 = 12 > 0

y( 3 ) "= 12 – 3∙3 2 = –15 < 0

At the point x = 2, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

Answer: 2

*For the same function, the minimum point is the point x = – 2.

77439. Find the maximum point of the function y = 9x 2 – x 3.

Let's find the derivative of the function:

Let's find the zeros of the derivative:

18x –3x 2 = 0

3x(6 – x) = 0

Solving the equation we get:

We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:

y(–1 ) "= 18 (–1) –3 (–1) 2 = –21< 0

y(1) "= 18∙1 –3∙1 2 = 15 > 0

y(7) "= 18∙7 –3∙7 2 = –1< 0

At the point x = 6, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

Answer: 6

*For the same function, the minimum point is the point x = 0.

77443. Find the maximum point of the function y = (x 3 /3)–9x–7.

Let's find the derivative of the function:

Let's find the zeros of the derivative:

x 2 – 9 = 0

x 2 = 9

Solving the equation we get:

We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:

y(–4 ) "= (–4) 2 – 9 > 0

y(0) "= 0 2 – 9 < 0

y(4) "= 4 2 – 9 > 0

At the point x = – 3, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

Answer: – 3

What is an extremum of a function and what is the necessary condition for an extremum?

The extremum of a function is the maximum and minimum of the function.

The necessary condition for the maximum and minimum (extremum) of a function is the following: if the function f(x) has an extremum at the point x = a, then at this point the derivative is either zero, or infinite, or does not exist.

This condition is necessary, but not sufficient. The derivative at the point x = a can go to zero, infinity, or not exist without the function having an extremum at this point.

What is the sufficient condition for the extremum of a function (maximum or minimum)?

First condition:

If, in sufficient proximity to the point x = a, the derivative f?(x) is positive to the left of a and negative to the right of a, then at the point x = a the function f(x) has maximum

If, in sufficient proximity to the point x = a, the derivative f?(x) is negative to the left of a and positive to the right of a, then at the point x = a the function f(x) has minimum provided that the function f(x) here is continuous.

Instead, you can use the second sufficient condition for the extremum of a function:

Let at the point x = a the first derivative f?(x) vanish; if the second derivative f??(a) is negative, then the function f(x) has a maximum at the point x = a, if it is positive, then it has a minimum.

What is the critical point of a function and how to find it?

This is the value of the function argument at which the function has an extremum (i.e. maximum or minimum). To find it you need find the derivative function f?(x) and, equating it to zero, solve the equation f?(x) = 0. The roots of this equation, as well as those points at which the derivative of this function does not exist, are critical points, i.e., values ​​of the argument at which there can be an extremum. They can be easily identified by looking at derivative graph: we are interested in those values ​​of the argument at which the graph of the function intersects the abscissa axis (Ox axis) and those at which the graph suffers discontinuities.

For example, let's find extremum of a parabola.

Function y(x) = 3x2 + 2x - 50.

Derivative of the function: y?(x) = 6x + 2

Solve the equation: y?(x) = 0

6x + 2 = 0, 6x = -2, x = -2/6 = -1/3

In this case, the critical point is x0=-1/3. It is with this argument value that the function has extremum. To him find, substitute the found number in the expression for the function instead of “x”:

y0 = 3*(-1/3)2 + 2*(-1/3) - 50 = 3*1/9 - 2/3 - 50 = 1/3 - 2/3 - 50 = -1/3 - 50 = -50.333.

How to determine the maximum and minimum of a function, i.e. its largest and smallest values?

If the sign of the derivative when passing through the critical point x0 changes from “plus” to “minus”, then x0 is maximum point; if the sign of the derivative changes from minus to plus, then x0 is minimum point; if the sign does not change, then at point x0 there is neither a maximum nor a minimum.

For the example considered:

We take an arbitrary value of the argument to the left of the critical point: x = -1

At x = -1, the value of the derivative will be y?(-1) = 6*(-1) + 2 = -6 + 2 = -4 (i.e. the sign is “minus”).

Now we take an arbitrary value of the argument to the right of the critical point: x = 1

At x = 1, the value of the derivative will be y(1) = 6*1 + 2 = 6 + 2 = 8 (i.e. the sign is “plus”).

As you can see, the derivative changed sign from minus to plus when passing through the critical point. This means that at the critical value x0 we have a minimum point.

Largest and smallest value of a function on the interval(on a segment) are found using the same procedure, only taking into account the fact that, perhaps, not all critical points will lie within the specified interval. Those critical points that are outside the interval must be excluded from consideration. If there is only one critical point within the interval, it will have either a maximum or a minimum. In this case, to determine the largest and smallest values ​​of the function, we also take into account the values ​​of the function at the ends of the interval.

For example, let's find the largest and smallest values ​​of the function

y(x) = 3sin(x) - 0.5x

at intervals:

So, the derivative of the function is

y?(x) = 3cos(x) - 0.5

We solve the equation 3cos(x) - 0.5 = 0

cos(x) = 0.5/3 = 0.16667

x = ±arccos(0.16667) + 2πk.

We find critical points on the interval [-9; 9]:

x = arccos(0.16667) - 2π*2 = -11.163 (not included in the interval)

x = -arccos(0.16667) - 2π*1 = -7.687

x = arccos(0.16667) - 2π*1 = -4.88

x = -arccos(0.16667) + 2π*0 = -1.403

x = arccos(0.16667) + 2π*0 = 1.403

x = -arccos(0.16667) + 2π*1 = 4.88

x = arccos(0.16667) + 2π*1 = 7.687

x = -arccos(0.16667) + 2π*2 = 11.163 (not included in the interval)

We find the values ​​of the function at critical values ​​of the argument:

y(-7.687) = 3cos(-7.687) - 0.5 = 0.885

y(-4.88) = 3cos(-4.88) - 0.5 = 5.398

y(-1.403) = 3cos(-1.403) - 0.5 = -2.256

y(1.403) = 3cos(1.403) - 0.5 = 2.256

y(4.88) = 3cos(4.88) - 0.5 = -5.398

y(7.687) = 3cos(7.687) - 0.5 = -0.885

It can be seen that on the interval [-9; 9] the function has the greatest value at x = -4.88:

x = -4.88, y = 5.398,

and the smallest - at x = 4.88:

x = 4.88, y = -5.398.

On the interval [-6; -3] we have only one critical point: x = -4.88. The value of the function at x = -4.88 is equal to y = 5.398.

Find the value of the function at the ends of the interval:

y(-6) = 3cos(-6) - 0.5 = 3.838

y(-3) = 3cos(-3) - 0.5 = 1.077

On the interval [-6; -3] we have the greatest value of the function

y = 5.398 at x = -4.88

smallest value -

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the convex and concave sides?

To find all the inflection points of the line y = f(x), you need to find the second derivative, equate it to zero (solve the equation) and test all those values ​​of x for which the second derivative is zero, infinite or does not exist. If, when passing through one of these values, the second derivative changes sign, then the graph of the function has an inflection at this point. If it doesn’t change, then there is no bend.

The roots of the equation f? (x) = 0, as well as possible discontinuity points of the function and the second derivative, divide the domain of definition of the function into a number of intervals. The convexity on each of their intervals is determined by the sign of the second derivative. If the second derivative at a point on the interval under study is positive, then the line y = f(x) is concave upward, and if negative, then downward.

How to find the extrema of a function of two variables?

To find the extrema of the function f(x,y), differentiable in the domain of its specification, you need:

1) find the critical points, and for this - solve the system of equations

fх? (x,y) = 0, fу? (x,y) = 0

2) for each critical point P0(a;b) investigate whether the sign of the difference remains unchanged

for all points (x;y) sufficiently close to P0. If the difference remains positive, then at point P0 we have a minimum, if negative, then we have a maximum. If the difference does not retain its sign, then there is no extremum at point P0.

The extrema of a function are determined similarly for a larger number of arguments.

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