The largest and smallest values ​​of the function

concepts of mathematical analysis. The value taken by a function at some point of the set on which this function is defined is called the largest (smallest) value on this set if the function does not have a larger (smaller) value at any other point in the set. N. and n. h. f. in comparison with its values ​​at all sufficiently close points are called extrema (respectively, maxima and minima) of the function. N. and n. h. f., given on a segment, can be achieved either at points where the derivative is equal to zero, or at points where it does not exist, or at the ends of the segment. A continuous function given on a segment necessarily reaches its maximum and minimum values; if a continuous function is considered on an interval (that is, a segment with excluded ends), then among its values ​​on this interval there may not be a maximum or a minimum. For example, the function at = x, given on the interval , reaches the largest and smallest values, respectively, at x= 1 and x= 0 (i.e., at the ends of the segment); if we consider this function on the interval (0; 1), then among its values ​​on this interval there is neither the largest nor the smallest, since for each x0 there is always a point of this interval lying to the right (to the left) x0, and such that the value of the function at this point will be greater (respectively, less) than at the point x0. Similar statements are valid for functions of several variables. See also Extreme.

Great Soviet Encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .

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Concepts of mathematical analysis. The value taken by the function at some point of the set on which this function is given is called the largest (smallest) on this set, if at no other point the function has a larger (smaller) ... ... encyclopedic Dictionary

The concepts of mathematics. analysis. The value taken by the function at a particular point of the set, pa rum this function is given, called. largest (smallest) on this set, if at no other point does the function have a larger (smaller) value ... Natural science. encyclopedic Dictionary

MAXIMUM AND MINIMUM FUNCTION- respectively, the largest and smallest values ​​of the function in comparison with its values ​​at all sufficiently close points. High and low points are called extreme points... Great Polytechnic Encyclopedia

The largest and, accordingly, the smallest values ​​of a function that takes real values. The point of the domain of definition of the function in question, in which it takes a maximum or minimum, is called. respectively the maximum point or the minimum point ... ... Mathematical Encyclopedia

A ternary function in the theory of functional systems and ternary logic is a function of type, where is a ternary set, and is a non-negative integer, which is called the arity or locality of the function. The elements of the set are digital ... ... Wikipedia

Representation of Boolean functions by normal forms (see Normal forms of Boolean functions). the simplest with respect to some measure of complexity. Usually, the complexity of a normal form is understood as the number of letters in it. In this case, the simplest form is called ... ... Mathematical Encyclopedia

A function that receives infinitesimal increments as the argument increments infinitesimally. A single-valued function f (x) is called continuous for the value of the argument x0, if for all values ​​of the argument x that differ sufficiently little from x0 ... Great Soviet Encyclopedia

- (Latin maximum and minimum, literally the largest and smallest) (Math.), the largest and smallest values ​​​​of a function compared to its values ​​\u200b\u200bin sufficiently close points. In the figure, the function y \u003d f (x) has a maximum at the points x1 and x3, and at the point x2 ... ... encyclopedic Dictionary

- (from the Latin maximum and minimum, the largest and smallest) (mathematical), the largest and smallest values ​​​​of a function compared to its values ​​\u200b\u200bin sufficiently close points. High and low points are called extreme points... Modern Encyclopedia

Sometimes in problems B15 there are "bad" functions for which it is difficult to find the derivative. Previously, this was only on probes, but now these tasks are so common that they can no longer be ignored when preparing for this exam.

In this case, other tricks work, one of which is - monotone.

The function f (x) is called monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following is true:

x 1< x 2 ⇒ f (x 1) < f (x2).

The function f (x) is called monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following is true:

x 1< x 2 ⇒ f (x 1) > f( x2).

In other words, for an increasing function, the larger x is, the larger f(x) is. For a decreasing function, the opposite is true: the more x , the smaller f(x).

For example, the logarithm increases monotonically if the base a > 1 and decreases monotonically if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.

f (x) = log a x (a > 0; a ≠ 1; x > 0)

The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:

The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, показательная функция определена для всех чисел, а не только для x > 0:

f (x) = a x (a > 0)

Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where monotony is broken.

All these functions are never found in their pure form. Polynomials, fractions and other nonsense are added to them, because of which it becomes difficult to calculate the derivative. What happens in this case - now we will analyze.

Parabola vertex coordinates

Most often, the function argument is replaced with square trinomial of the form y = ax 2 + bx + c . Its graph is a standard parabola, in which we are interested in:

1. Parabola branches - can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
2. The vertex of a parabola is the extremum point of a quadratic function, at which this function takes its smallest (for a > 0) or largest (a< 0) значение.

Of greatest interest is top of a parabola, the abscissa of which is calculated by the formula:

So, we have found the extremum point of the quadratic function. But if the original function is monotonic, for it the point x 0 will also be an extremum point. Thus, we formulate the key rule:

The extremum points of the square trinomial and the complex function it enters into coincide. Therefore, you can look for x 0 for a square trinomial, and forget about the function.

From the above reasoning, it remains unclear what kind of point we get: a maximum or a minimum. However, the tasks are specifically designed so that it does not matter. Judge for yourself:

1. There is no segment in the condition of the problem. Therefore, it is not required to calculate f(a) and f(b). It remains to consider only the extremum points;
2. But there is only one such point - this is the top of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.

Thus, the solution of the problem is greatly simplified and reduced to just two steps:

1. Write out the parabola equation y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a;
2. Find the value of the original function at this point: f (x 0). If there are no additional conditions, this will be the answer.

At first glance, this algorithm and its justification may seem complicated. I deliberately do not post a "bare" solution scheme, since the thoughtless application of such rules is fraught with errors.

Consider the real tasks from the trial exam in mathematics - this is where this technique is most common. At the same time, we will make sure that in this way many problems of B15 become almost verbal.

Under the root is a quadratic function y \u003d x 2 + 6x + 13. The graph of this function is a parabola with branches up, since the coefficient a \u003d 1\u003e 0.

Top of the parabola:

x 0 \u003d -b / (2a) \u003d -6 / (2 1) \u003d -6 / 2 \u003d -3

Since the branches of the parabola are directed upwards, at the point x 0 \u003d −3, the function y \u003d x 2 + 6x + 13 takes on the smallest value.

The root is monotonically increasing, so x 0 is the minimum point of the entire function. We have:

Task. Find the smallest value of the function:

y = log 2 (x 2 + 2x + 9)

Under the logarithm is again a quadratic function: y \u003d x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.

Top of the parabola:

x 0 \u003d -b / (2a) \u003d -2 / (2 1) \u003d -2/2 \u003d -1

So, at the point x 0 = −1, the quadratic function takes on the smallest value. But the function y = log 2 x is monotone, so:

y min = y (−1) = log 2 ((−1) 2 + 2 (−1) + 9) = ... = log 2 8 = 3

The exponent is a quadratic function y = 1 − 4x − x 2 . Let's rewrite it in normal form: y = −x 2 − 4x + 1.

Obviously, the graph of this function is a parabola, branches down (a = −1< 0). Поэтому вершина будет точкой максимума:

x 0 = −b /(2a ) = −(−4)/(2 (−1)) = 4/(−2) = −2

The original function is exponential, it is monotone, so the largest value will be at the found point x 0 = −2:

An attentive reader will surely notice that we did not write out the area of ​​\u200b\u200bpermissible values ​​of the root and logarithm. But this was not required: inside there are functions whose values ​​are always positive.

Consequences from the scope of a function

Sometimes, to solve problem B15, it is not enough just to find the vertex of the parabola. The desired value may lie at the end of the segment, but not at the extremum point. If the task does not specify a segment at all, look at tolerance range original function. Namely:

Pay attention again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how it works with specific examples:

Task. Find the largest value of the function:

Under the root is again a quadratic function: y \u003d 3 - 2x - x 2. Its graph is a parabola, but branches down since a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический квадратный корень из отрицательного числа не существует.

We write out the area of ​​​​permissible values ​​​​(ODZ):

3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; one]

Now find the vertex of the parabola:

x 0 = −b /(2a ) = −(−2)/(2 (−1)) = 2/(−2) = −1

The point x 0 = −1 belongs to the ODZ segment - and this is good. Now we consider the value of the function at the point x 0, as well as at the ends of the ODZ:

y(−3) = y(1) = 0

So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.

Task. Find the smallest value of the function:

y = log 0.5 (6x - x 2 - 5)

Inside the logarithm there is a quadratic function y \u003d 6x - x 2 - 5. This is a parabola with branches down, but there cannot be negative numbers in the logarithm, so we write out the ODZ:

6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)

Please note: the inequality is strict, so the ends do not belong to the ODZ. In this way, the logarithm differs from the root, where the ends of the segment suit us quite well.

Looking for the vertex of the parabola:

x 0 = −b /(2a ) = −6/(2 (−1)) = −6/(−2) = 3

The top of the parabola fits along the ODZ: x 0 = 3 ∈ (1; 5). But since the ends of the segment do not interest us, we consider the value of the function only at the point x 0:

y min = y (3) = log 0.5 (6 3 − 3 2 − 5) = log 0.5 (18 − 9 − 5) = log 0.5 4 = −2

The largest and smallest value of the function

The largest value of a function is called the largest, the smallest value is the smallest of all its values.

A function may have only one largest and only one smallest value, or may not have any at all. Finding the largest and smallest values ​​of continuous functions is based on the following properties of these functions:

1) If in some interval (finite or infinite) the function y=f(x) is continuous and has only one extremum, and if this is the maximum (minimum), then it will be the largest (smallest) value of the function in this interval.

2) If the function f(x) is continuous on a certain segment, then it necessarily has the largest and smallest values ​​on this segment. These values ​​are reached either at the extremum points lying inside the segment, or at the boundaries of this segment.

To find the largest and smallest values ​​on the segment, it is recommended to use the following scheme:

1. Find the derivative.

2. Find the critical points of the function where =0 or does not exist.

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

When solving applied problems, in particular optimization problems, the problems of finding the largest and smallest values ​​(global maximum and global minimum) of a function on the interval X are important. To solve such problems, one should, based on the condition, choose an independent variable and express the value under study through this variable. Then find the desired maximum or minimum value of the resulting function. In this case, the interval of change of the independent variable, which can be finite or infinite, is also determined from the condition of the problem.

Example. The tank, which has the shape of a rectangular parallelepiped with a square bottom, open at the top, must be tinned inside with tin. What should be the dimensions of the tank with a capacity of 108 liters. water so that the cost of its tinning is the least?

Decision. The cost of coating the tank with tin will be the lowest if, for a given capacity, its surface is minimal. Denote by a dm - the side of the base, b dm - the height of the tank. Then the area S of its surface is equal to

And

The resulting relation establishes the relationship between the surface area of ​​the tank S (function) and the side of the base a (argument). We investigate the function S for an extremum. Find the first derivative, equate it to zero and solve the resulting equation:

Hence a = 6. (a) > 0 for a > 6, (a)< 0 при а < 6. Следовательно, при а = 6 функция S имеет минимум. Если а = 6, то b = 3. Таким образом, затраты на лужение резервуара емкостью 108 литров будут наименьшими, если он имеет размеры 6дм х 6дм х 3дм.

Example. Find the largest and smallest values ​​of a function in between.

Decision: The specified function is continuous on the entire number axis. Function derivative

Derivative at and at . Let's calculate the values ​​of the function at these points:

.

The function values ​​at the ends of the given interval are equal to . Therefore, the largest value of the function is at , the smallest value of the function is at .

Questions for self-examination

1. Formulate L'Hopital's rule for disclosure of uncertainties of the form . List the different types of uncertainties for which L'Hospital's rule can be used.

2. Formulate signs of increasing and decreasing function.

3. Define the maximum and minimum of a function.

4. Formulate the necessary condition for the existence of an extremum.

5. What values ​​of the argument (what points) are called critical? How to find these points?

6. What are sufficient signs of the existence of an extremum of a function? Outline a scheme for studying a function for an extremum using the first derivative.

7. Outline the scheme for studying the function for an extremum using the second derivative.

8. Define convexity, concavity of a curve.

9. What is the inflection point of a function graph? Specify how to find these points.

10. Formulate the necessary and sufficient signs of convexity and concavity of the curve on a given segment.

11. Define the asymptote of the curve. How to find the vertical, horizontal and oblique asymptotes of a function graph?

12. Outline the general scheme for researching a function and constructing its graph.

13. Formulate a rule for finding the largest and smallest values ​​of a function on a given interval.

How to find the largest and smallest values ​​of a function on a segment?

For this we follow the well-known algorithm:

1 . We find ODZ functions.

2 . Finding the derivative of a function

3 . Equate the derivative to zero

4 . We find the intervals at which the derivative retains its sign, and from them we determine the intervals of increase and decrease of the function:

If on the interval I the derivative of the function 0" title="(!LANG:f^(prime)(x)>0">, то функция !} increases over this interval.

If on the interval I the derivative of the function , then the function decreases over this interval.

5 . We find maximum and minimum points of the function.

AT the function maximum point, the derivative changes sign from "+" to "-".

AT minimum point of the functionderivative changes sign from "-" to "+".

6 . We find the value of the function at the ends of the segment,

• then we compare the value of the function at the ends of the segment and at the maximum points, and choose the largest of them if you need to find the largest value of the function
• or we compare the value of the function at the ends of the segment and at the minimum points, and choose the smallest of them if you need to find the smallest value of the function

However, depending on how the function behaves on the interval, this algorithm can be significantly reduced.

Consider the function . The graph of this function looks like this:

Let's consider several examples of solving problems from the Open Task Bank for

one . Task B15 (#26695)

On the cut.

1. The function is defined for all real values ​​of x

Obviously, this equation has no solutions, and the derivative is positive for all values ​​of x. Therefore, the function increases and takes on the largest value at the right end of the interval, that is, at x=0.

Answer: 5.

2 . Task B15 (No. 26702)

Find the largest value of a function on the segment.

1.ODZ function title="(!LANG:x(pi)/2+(pi)k, k(in)(bbZ)">!}

The derivative is zero at , however, at these points it does not change sign:

Therefore, title="(!LANG:3/(cos^2(x))>=3">, значит, title="3/(cos^2(x))-3>=0">, то есть производная при всех допустимых значених х неотрицательна, следовательно, функция !} increases and takes the greatest value at the right end of the interval, at .

To make it clear why the derivative does not change sign, we transform the expression for the derivative as follows:

Title="(!LANG:y^(prime)=3/(cos^2(x))-3=(3-3cos^2(x))/(cos^2(x))=(3sin^2 (x))/(cos^2(x))=3tg^2(x)>=0">!}

Answer: 5.

3 . Task B15 (#26708)

Find the smallest value of the function on the interval .

1. ODZ functions: title="(!LANG:x(pi)/2+(pi)k, k(in)(bbZ)">!}

Let's place the roots of this equation on a trigonometric circle.

The interval contains two numbers: and

Let's put up the signs. To do this, we determine the sign of the derivative at the point x=0: . When passing through the points and the derivative changes sign.

Let's depict the change of signs of the derivative of the function on the coordinate line:

Obviously, the point is a minimum point (where the derivative changes sign from "-" to "+"), and in order to find the smallest value of the function on the segment, you need to compare the function values ​​at the minimum point and at the left end of the segment, .

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 listed lines (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 what this surface looks like 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 in 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 in 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

In a limited enclosed area

Decision: 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 the 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) To study the right side of the triangle, we substitute it into the function and “put things in order there”:

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

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 appeared point:

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 unusual examples for you 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.

Pay special attention to the rational order and technique of studying the border 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 bold 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

The final examples are devoted to other useful ideas that will come in handy 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 ;-)

Decision, 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:

The 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 have read these lines =) If in the two previous examples calculations in decimal fractions were convenient (which, by the way, is rare), then here we are waiting for the usual ordinary 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 a standalone solution:

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: Decision: draw the area on the drawing: