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The largest and smallest value of the function definition is short. The largest and smallest values ​​of a function on a segment

The study of such an object of mathematical analysis as a function is of great importance. meaning and in other areas of science. For example, in economic analysis constantly need to evaluate behavior functions profit, namely to determine its maximum meaning and develop a strategy to achieve it.

Instruction

The study of any behavior should always begin with a search for a domain of definition. Usually, according to the condition of a particular problem, it is required to determine the largest meaning functions either on the whole of this area, or on its specific interval with open or closed boundaries.

Based on , the largest is meaning functions y(x0), under which for any point of the domain of definition the inequality y(x0) ≥ y(x) (х ≠ x0) is satisfied. Graphically, this point will be the highest if you arrange the values ​​of the argument along the abscissa axis, and the function itself along the ordinate axis.

To determine the largest meaning functions, follow the three-step algorithm. Note that you must be able to work with one-sided and , as well as calculate the derivative. So, let some function y(x) be given and it is required to find its largest meaning on some interval with boundary values ​​A and B.

Find out if this interval is within the scope functions. To do this, you need to find it, considering all possible restrictions: the presence of a fraction in the expression, square root etc. The domain of definition is the set of argument values ​​for which the function makes sense. Determine if the given interval is a subset of it. If yes, then proceed to the next step.

Find the derivative functions and solve the resulting equation by equating the derivative to zero. Thus, you will get the values ​​of the so-called stationary points. Evaluate if at least one of them belongs to the interval A, B.

Consider these points at the third stage, substitute their values ​​into the function. Perform the following additional steps depending on the interval type. If there is a segment of the form [A, B], the boundary points are included in the interval, this is indicated by brackets. Calculate Values functions for x = A and x = B. If the open interval is (A, B), the boundary values ​​are punctured, i.e. are not included in it. Solve one-sided limits for x→A and x→B. A combined interval of the form [A, B) or (A, B), one of whose boundaries belongs to it, the other does not. Find the one-sided limit as x tends to the punctured value, and substitute the other into the function. Infinite two-sided interval (-∞, +∞) or one-sided infinite intervals of the form: , (-∞, B) For real limits A and B, proceed according to the principles already described, and for infinite, look for limits for x→-∞ and x→+∞, respectively.

The task at this stage

Given a function that is defined and continuous on some interval . It is required to find the largest (smallest) value of the function on this interval.

Theoretical basis.
Theorem (Second Weierstrass Theorem):

If a function is defined and continuous in a closed interval , then it reaches its maximum and minimum values ​​in this interval.

The function can reach its maximum and minimum values ​​either at the internal points of the interval or at its boundaries. Let's illustrate all possible options.

Explanation:
1) The function reaches its maximum value on the left border of the interval at the point , and its minimum value on the right border of the interval at the point .
2) The function reaches its maximum value at the point (this is the maximum point), and its minimum value at the right boundary of the interval at the point.
3) The function reaches its maximum value on the left border of the interval at the point , and its minimum value at the point (this is the minimum point).
4) The function is constant on the interval, i.e. it reaches its minimum and maximum values ​​at any point in the interval, and the minimum and maximum values ​​are equal to each other.
5) The function reaches its maximum value at the point , and its minimum value at the point (despite the fact that the function has both a maximum and a minimum on this interval).
6) The function reaches its maximum value at a point (this is the maximum point), and its minimum value at a point (this is the minimum point).
Comment:

"Maximum" and "maximum value" are different things. This follows from the definition of the maximum and the intuitive understanding of the phrase "maximum value".

Algorithm for solving problem 2.



4) Choose the largest (smallest) from the obtained values ​​and write down the answer.

Example 4:

Determine the largest and smallest value functions on the segment.
Solution:
1) Find the derivative of the function.

2) Find stationary points (and points that are suspicious of an extremum) by solving the equation . Pay attention to the points where there is no two-sided finite derivative.

3) Calculate the values ​​of the function at stationary points and at the boundaries of the interval.

4) Choose the largest (smallest) from the obtained values ​​and write down the answer.

The function on this segment reaches its maximum value at the point with coordinates .

The function on this segment reaches its minimum value at the point with coordinates .

You can verify the correctness of the calculations by looking at the graph of the function under study.


Comment: The function reaches its maximum value at the maximum point, and the minimum value at the boundary of the segment.

Special case.

Suppose you want to find the maximum and minimum value of some function on a segment. After the execution of the first paragraph of the algorithm, i.e. calculation of the derivative, it becomes clear that, for example, it takes only negative values ​​on the entire segment under consideration. Remember that if the derivative is negative, then the function is decreasing. We found that the function is decreasing on the entire interval. This situation is shown in the chart No. 1 at the beginning of the article.

The function decreases on the interval, i.e. it has no extremum points. It can be seen from the picture that the function will take the smallest value on the right border of the segment, and the largest value on the left. if the derivative on the interval is everywhere positive, then the function is increasing. The smallest value is on the left border of the segment, the largest is on the right.

And to solve it, you need minimal knowledge of the topic. The next academic year is ending, everyone wants to go on vacation, and in order to bring this moment closer, I immediately get down to business:

Let's start with the area. The area referred to in the condition is limited closed set of points in the plane. For example, a set of points bounded by a triangle, including the ENTIRE triangle (if from borders“Poke out” at least one point, then the area will no longer be closed). In practice, there are also areas of rectangular, round and slightly more complex shapes. It should be noted that in the theory of mathematical analysis, strict definitions are given limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and more is not needed now.

The flat area is standardly denoted by the letter , and, as a rule, is given analytically - by several equations (not necessarily linear); less often inequalities. A typical verbal turnover: "closed arealimited by lines".

An integral part of the task under consideration is the construction of the area on the drawing. How to do it? It is necessary to draw all the lines listed (in this case 3 straight) and analyze what happened. The desired area is usually lightly hatched, and its border is highlighted with a bold line:


The same area can be set linear inequalities: , which for some reason are more often written as an enumeration list, and not system.
Since the boundary belongs to the region, then all inequalities, of course, non-strict.

And now the crux of the matter. Imagine that the axis goes straight to you from the origin of coordinates. Consider a function that continuous in each area point. The graph of this function is surface, and the small happiness is that in order to solve today's problem, we do not need to know how this surface looks at all. It can be located above, below, cross the plane - all this is not important. And the following is important: according to Weierstrass theorems, continuous V limited closed area, the function reaches its maximum (of the "highest") and least (of the "lowest") values ​​to be found. These values ​​are achieved or V stationary points, belonging to the regionD , or at points that lie on the boundary of this region. From which follows a simple and transparent solution algorithm:

Example 1

Limited closed area

Solution: First of all, you need to depict the area on the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the problem, and therefore I will immediately give the final illustration, which shows all the "suspicious" points found during the study. Usually they are put down one after another as they are found:

Based on the preamble, the decision can be conveniently divided into two points:

I) Let's find stationary points. This is a standard action that we have repeatedly performed in the lesson. about extrema of several variables:

Found stationary point belongs areas: (mark it on the drawing), which means that we should calculate the value of the function at a given point:

- as in the article The largest and smallest values ​​of a function on a segment, I will highlight the important results in bold. In a notebook, it is convenient to circle them with a pencil.

Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at the point the function reaches, for example, local minimum, then this DOES NOT MEAN that the resulting value will be minimal throughout the region (see the beginning of the lesson about unconditional extremes) .

What if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and go to the next paragraph.

II) We investigate the border of the region.

Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subparagraphs. But it is better to do it not anyhow. From my point of view, at first it is more advantageous to consider segments parallel to the coordinate axes, and first of all, those lying on the axes themselves. To catch the whole sequence and logic of actions, try to study the ending "in one breath":

1) Let's deal with the lower side of the triangle. To do this, we substitute directly into the function:

Alternatively, you can do it like this:

Geometrically, this means that the coordinate plane (which is also given by the equation)"cut out" from surfaces"spatial" parabola, the top of which immediately falls under suspicion. Let's find out where is she:

- the resulting value "hit" in the area, and it may well be that at the point (mark on the drawing) the function reaches the largest or smallest value in the entire area. Anyway, let's do the calculations:

Other "candidates" are, of course, the ends of the segment. Calculate the values ​​of the function at points (mark on the drawing):

Here, by the way, you can perform an oral mini-check on the "stripped down" version:

2) For research right side we substitute the triangle into the function and “put things in order there”:

Here we immediately perform a rough check, “ringing” the already processed end of the segment:
, Great.

The geometric situation is related to the previous point:

- the resulting value also “entered the scope of our interests”, which means that we need to calculate what the function is equal to at the point that has appeared:

Let's examine the second end of the segment:

Using the function , let's check:

3) Everyone probably knows how to explore the remaining side. We substitute into the function and carry out simplifications:

Line ends have already been investigated, but on the draft we still check whether we found the function correctly :
– coincided with the result of the 1st subparagraph;
– coincided with the result of the 2nd subparagraph.

It remains to find out if there is something interesting inside the segment :

- There is! Substituting a straight line into the equation, we get the ordinate of this “interestingness”:

We mark a point on the drawing and find the corresponding value of the function:

Let's control the calculations according to the "budget" version :
, order.

And the final step: CAREFULLY look through all the "fat" numbers, I recommend even beginners to make a single list:

from which we choose the largest and smallest values. Answer write in the style of the problem of finding the largest and smallest values ​​of the function on the interval:

Just in case, I will once again comment on the geometric meaning of the result:
– here is the highest point of the surface in the region ;
- here is the lowest point of the surface in the area.

In the analyzed problem, we found 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum "exploration set" consists of three points. This happens when the function, for example, sets plane- it is quite clear that there are no stationary points, and the function can reach the maximum / minimum values ​​only at the vertices of the triangle. But there are no such examples once, twice - usually you have to deal with some kind of surface of the 2nd order.

If you solve such tasks a little, then triangles can make your head spin, and therefore I have prepared for you unusual examples to make it square :)

Example 2

Find the largest and smallest values ​​of a function in a closed area bounded by lines

Example 3

Find the largest and smallest values ​​of a function in a bounded closed area.

Special attention pay attention to the rational order and technique of studying the boundary of the area, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it as you like, but in some problems, for example, in the same Example 2, there is every chance to significantly complicate your life. An approximate example of finishing assignments at the end of the lesson.

We systematize the solution algorithm, otherwise, with my diligence of a spider, it somehow got lost in a long thread of comments of the 1st example:

- At the first step, we build an area, it is desirable to shade it, and highlight the border with a thick line. During the solution, points will appear that need to be put on the drawing.

– Find stationary points and calculate the values ​​of the function only in those, which belong to the area . The obtained values ​​are highlighted in the text (for example, circled with a pencil). If the stationary point does NOT belong to the area, then we mark this fact with an icon or verbally. If there are no stationary points at all, then we draw a written conclusion that they are absent. In any case, this item cannot be skipped!

– Exploring the border area. First, it is advantageous to deal with straight lines that are parallel to the coordinate axes (if there are any). The function values ​​calculated at "suspicious" points are also highlighted. A lot has been said about the solution technique above and something else will be said below - read, re-read, delve into!

- From the selected numbers, select the largest and smallest values ​​\u200b\u200band give an answer. Sometimes it happens that the function reaches such values ​​at several points at once - in this case, all these points should be reflected in the answer. Let, for example, and it turned out that this is the smallest value. Then we write that

Final examples are dedicated to others useful ideas useful in practice:

Example 4

Find the largest and smallest values ​​of a function in a closed area .

I have kept the author's formulation, in which the area is given as a double inequality. This condition can be written in an equivalent system or in a more traditional form for this problem:

I remind you that with non-linear we encountered inequalities on , and if you do not understand the geometric meaning of the entry, then please do not delay and clarify the situation right now ;-)

Solution, as always, begins with the construction of the area, which is a kind of "sole":

Hmm, sometimes you have to gnaw not only the granite of science ....

I) Find stationary points:

Idiot's dream system :)

The stationary point belongs to the region, namely, lies on its boundary.

And so, it’s nothing ... fun lesson went - that’s what it means to drink the right tea =)

II) We investigate the border of the region. Without further ado, let's start with the x-axis:

1) If , then

Find where the top of the parabola is:
- Appreciate such moments - "hit" right to the point, from which everything is already clear. But don't forget to check:

Let's calculate the values ​​of the function at the ends of the segment:

2) We will deal with the lower part of the “sole” “in one sitting” - without any complexes we substitute it into the function, moreover, we will only be interested in the segment:

Control:

Now this is already bringing some revival to the monotonous ride on a knurled track. Let's find the critical points:

We decide quadratic equation do you remember this one? ... However, remember, of course, otherwise you would not read these lines =) If in the two previous examples calculations were convenient in decimal fractions(which, by the way, is rare), then here we are waiting for the usual common fractions. We find the “x” roots and, using the equation, determine the corresponding “game” coordinates of the “candidate” points:


Let's calculate the values ​​of the function at the found points:

Check the function yourself.

Now we carefully study the won trophies and write down answer:

Here are the "candidates", so the "candidates"!

For independent decision:

Example 5

Find the smallest and largest values ​​of a function in a closed area

An entry with curly braces reads like this: “a set of points such that”.

Sometimes in such examples they use Lagrange multiplier method, but the real need to use it is unlikely to arise. So, for example, if a function with the same area "de" is given, then after substitution into it - with a derivative of no difficulties; moreover, everything is drawn up in a “one line” (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are more complicated cases, where without the Lagrange function (where , for example, is the same circle equation) it's hard to get by - how hard it is to get by without a good rest!

All the best to pass the session and see you soon next season!

Solutions and answers:

Example 2: Solution: draw the area on the drawing:

The process of finding the smallest and largest values ​​of a function on a segment is reminiscent of a fascinating flight around an object (a graph of a function) on a helicopter with firing from a long-range cannon at certain points and choosing from these points very special points for control shots. Points are selected in a certain way and according to certain rules. By what rules? We will talk about this further.

If the function y = f(x) continuous on the interval [ a, b] , then it reaches on this segment least And highest values . This can either happen in extremum points or at the ends of the segment. Therefore, to find least And the largest values ​​of the function , continuous on the interval [ a, b] , you need to calculate its values ​​in all critical points and at the ends of the segment, and then choose the smallest and largest of them.

Let, for example, it is required to determine the maximum value of the function f(x) on the segment [ a, b] . To do this, find all its critical points lying on [ a, b] .

critical point is called the point at which function defined, and her derivative is either zero or does not exist. Then you should calculate the values ​​of the function at critical points. And, finally, one should compare the values ​​of the function at critical points and at the ends of the segment ( f(a) And f(b) ). The largest of these numbers will be the largest value of the function on the segment [a, b] .

The problem of finding the smallest values ​​of the function .

We are looking for the smallest and largest values ​​​​of the function together

Example 1. Find the smallest and largest values ​​of a function on the segment [-1, 2] .

Solution. We find the derivative of this function. Equate the derivative to zero () and get two critical points: and . To find the smallest and largest values ​​of a function on a given segment, it is enough to calculate its values ​​at the ends of the segment and at the point , since the point does not belong to the segment [-1, 2] . These function values ​​are the following: , , . It follows that smallest function value(marked in red on the graph below), equal to -7, is reached at the right end of the segment - at the point , and greatest(also red on the graph), is equal to 9, - at the critical point .

If the function is continuous in a certain interval and this interval is not a segment (but is, for example, an interval; the difference between an interval and a segment: the boundary points of the interval are not included in the interval, but the boundary points of the segment are included in the segment), then among the values ​​of the function there may not be be the smallest and largest. So, for example, the function depicted in the figure below is continuous on ]-∞, +∞[ and does not have the largest value.

However, for any interval (closed, open, or infinite), the following property of continuous functions holds.

Example 4. Find the smallest and largest values ​​of a function on the segment [-1, 3] .

Solution. We find the derivative of this function as the derivative of the quotient:

.

We equate the derivative to zero, which gives us one critical point: . It belongs to the interval [-1, 3] . To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Let's compare these values. Conclusion: equal to -5/13, at the point and the greatest value equal to 1 at the point .

We continue to search for the smallest and largest values ​​​​of the function together

There are teachers who, on the topic of finding the smallest and largest values ​​of a function, do not give students examples more complicated than those just considered, that is, those in which the function is a polynomial or a fraction, the numerator and denominator of which are polynomials. But we will not limit ourselves to such examples, since among teachers there are lovers of making students think in full (table of derivatives). Therefore, the logarithm and the trigonometric function will be used.

Example 6. Find the smallest and largest values ​​of a function on the segment .

Solution. We find the derivative of this function as derivative of the product :

We equate the derivative to zero, which gives one critical point: . It belongs to the segment. To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

The result of all actions: the function reaches its minimum value, equal to 0, at a point and at a point and the greatest value equal to e² , at the point .

Example 7. Find the smallest and largest values ​​of a function on the segment .

Solution. We find the derivative of this function:

Equate the derivative to zero:

The only critical point belongs to the segment . To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Conclusion: the function reaches its minimum value, equal to , at the point and the greatest value, equal to , at the point .

In applied extremal problems, finding the smallest (largest) function values, as a rule, is reduced to finding the minimum (maximum). But it is not the minima or maxima themselves that are of greater practical interest, but the values ​​of the argument at which they are achieved. When solving applied problems, an additional difficulty arises - the compilation of functions that describe the phenomenon or process under consideration.

Example 8 A tank with a capacity of 4, having the shape of a parallelepiped with a square base and open at the top, must be tinned. What should be the dimensions of the tank in order to cover it with the least amount of material?

Solution. Let x- base side h- tank height, S- its surface area without cover, V- its volume. The surface area of ​​the tank is expressed by the formula , i.e. is a function of two variables. To express S as a function of one variable, we use the fact that , whence . Substituting the found expression h into the formula for S:

Let us examine this function for an extremum. It is defined and differentiable everywhere in ]0, +∞[ , and

.

We equate the derivative to zero () and find the critical point. In addition, at , the derivative does not exist, but this value is not included in the domain of definition and therefore cannot be an extremum point. So, - the only critical point. Let's check it for the presence of an extremum using the second sufficient sign. Let's find the second derivative. When the second derivative is greater than zero (). This means that when the function reaches a minimum . Because this minimum - the only extremum of this function, it is its smallest value. So, the side of the base of the tank should be equal to 2 m, and its height.

Example 9 From paragraph A, located on the railway line, to the point WITH, at a distance from it l, goods must be transported. The cost of transporting a weight unit per unit distance by rail is equal to , and by highway it is equal to . To what point M lines railway a highway should be built so that the transportation of goods from A V WITH was the most economical AB railroad is assumed to be straight)?

petite and pretty simple task from the category of those that serve as a lifeline for a floating student. In nature, the sleepy realm of mid-July, so it's time to settle down with a laptop on the beach. Early in the morning, a sunbeam of theory played to soon focus on practice, which, despite its declared lightness, contains glass fragments in the sand. In this regard, I recommend conscientiously consider a few examples of this page. To solve practical tasks, you need to be able to find derivatives and understand the material of the article Intervals of monotonicity and extrema of a function.

First, briefly about the main thing. In a lesson about function continuity I gave the definition of continuity at a point and continuity on an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on a segment if:

1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.

The second paragraph deals with the so-called unilateral continuity functions at a point. There are several approaches to its definition, but I will stick to the line started earlier:

The function is continuous at a point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: . It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at that point:

Imagine that the green dots are the nails on which the magic rubber band is attached:

Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited- a hedge above, a hedge below, and our product grazes in a paddock. Thus, a function continuous on a segment is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and rigorously proved Weierstrass' first theorem.… Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled the graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Indeed, how do you know what awaits us beyond the horizon? After all, once the Earth was considered flat, so today even ordinary teleportation requires proof =)

According to second Weierstrass theorem, continuous on the segmentfunction reaches its exact top edge and his exact bottom edge .

The number is also called the maximum value of the function on the segment and denoted by , and the number - the minimum value of the function on the interval marked .

In our case:

Note : in theory, records are common .

Roughly speaking, the largest value is located where the highest point of the graph, and the smallest - where the lowest point.

Important! As already pointed out in the article on extrema of the function, the largest value of the function And smallest function value – NOT THE SAME, What function maximum And function minimum. So, in this example, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even the flood, in the context of the problem under consideration, this does not interest us at all. The task involves only finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore, no need to draw!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the function values ​​in critical points, that belong to this segment.

Catch one more goodie: there is no need to check a sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum not yet guaranteed what is the minimum or maximum value. The demonstration function reaches its maximum and, by the will of fate, the same number is the largest value of the function on the interval . But, of course, such a coincidence does not always take place.

So, at the first step, it is faster and easier to calculate the values ​​of the function at critical points belonging to the segment, without bothering whether they have extrema or not.

2) We calculate the values ​​of the function at the ends of the segment.

3) Among the values ​​of the function found in the 1st and 2nd paragraphs, we select the smallest and most big number, write down the answer.

We sit on the shore of the blue sea and hit the heels in shallow water:

Example 1

Find the largest and smallest values ​​of a function on a segment

Solution:
1) Calculate the values ​​of the function at critical points belonging to this segment:

Let us calculate the value of the function at the second critical point:

2) Calculate the values ​​of the function at the ends of the segment:

3) "Bold" results were obtained with exponentials and logarithms, which significantly complicates their comparison. For this reason, we will arm ourselves with a calculator or Excel and calculate the approximate values, not forgetting that:

Now everything is clear.

Answer:

Fractional-rational instance for independent solution:

Example 6

Find the maximum and minimum values ​​of a function on a segment

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