The greatest value at the interval. The greatest and smallest values \u200b\u200bof the function on the segment. Tasks for finding the largest and smallest values \u200b\u200bof values

19.02.2021 Complications

The study of functions and their graphs is a topic that pays special attention within the framework of the school program of high schools. Some foundations of mathematical analysis - differentiation are included in the profile level of the exam in mathematics. Some schoolchildren have problems with this topic, as they confuse the graphics of the function and the derivative, as well as forget algorithms. This article will consider the main types of tasks and how to solve them.

What is the meaning of the function?

The mathematical function is a special equation. It establishes the relationship between numbers. The function depends on the value of the argument.

The value of the function is calculated according to a given formula. To do this, substitute any argument, which corresponds to the area of \u200b\u200bpermissible values, in this formula in place and perform the necessary mathematical operations. What kind?

How can I find the smallest function of the function using a function graph?

The graphic image of the dependency function from the argument is called a function graph. It is built on a plane with a specific unit segment, where the value of the variable, or the argument is deployed along the horizontal axis of the abscissa, and according to the vertical axis of the ordinate - the function corresponding to it.

The greater the value of the argument, the point itself it lies on the chart. And the larger the value of the function itself, the higher the point is.

What does it say about? The smallest value of the function will be the point that lies below the entire schedule. In order to find it on the segment of the graph, you need:

1) Find and note the ends of this segment.

2) Visually determine which point on this segment lies below all.

3) In response to write down its numeric value, which can be determined by sprocating the point on the ordinate axis.

on the derivative graph. Where to looking for?

However, when solving tasks, a schedule is sometimes given a function, but its derivative. In order to accidentally not allow a stupid error, it is better to carefully read the conditions, since it depends on it, where you need to look for extremum points.

So, the derivative is an instantaneous increase in the function. According to the geometrical definition, the derivative corresponds to the angular coefficient of tangential, which is directly carried out to this point.

It is known that at the extremum points tangency parallel to the OX axis. This means that its angular coefficient is 0.

From this we can conclude that at extremum points the derivative lies on the abscissa axis or turns into zero. But besides, at these points, the function changes its direction. That is, after a period of increasing, it starts to decrease, and the derivative, respectively, is replaced with a positive to negative. Or vice versa.

If a derivative of positive becomes negative - this is a maximum point. If the negative becomes positive - the minimum point.

Important: If the task requires a minimum or maximum point, then in response, the corresponding value along the abscissa axis should be written. But if you want to find the value of the function, then you first need to substitute the corresponding value of the argument to the function and calculate it.

How to find extremum points using a derivative?

The considered examples are mainly related to the task at number 7 of the exam, which implies work with a graph derivative or primitive. But the task is 12 EGE - to find the smallest value of the function on the segment (sometimes the largest) - performed without any drawings and requires basic skills of mathematical analysis.

To fulfill it, you need to be able to find the points of extremum using a derivative. The algorithm of their location is:

Find a derivative of the function.
Equate it to zero.
Find the roots of the equation.
Check whether the points obtained are extremum or inflection points.

To do this, it is necessary to draw the scheme and on the resulting intervals to determine the signs of the derivative, substituting the numbers belonging to the segments in the derivative. If, when solving the equation, you received the roots of double multiplicity - these are points of inflection.

Applying the theorems, determine which points are points of a minimum, and which is the maximum.

Calculation of the smallest value of the function using the derivative

However, by completing all these actions, we will find the values \u200b\u200bof the minimum points and the maximum along the abscissa axis. But how to find the smallest value of the function on the segment?

What needs to be done in order to find a function that matches the function at a specific point? It is necessary to substitute the value of the argument into this formula.

The minimum and maximum points correspond to the smallest and greatest value of the function on the segment. So, to find the value of the function, you need to calculate the function using the obtained x.

Important! If the task requires a minimum or maximum point, then in response to write the corresponding value along the abscissa axis. But if you need to find the value of the function, you must first substitute the corresponding value of the argument to the function and perform the necessary mathematical operations.

What if there are no minimum points on this segment?

But how to find the smallest value of the function on the segment on which there are no extremum points?

This means that on it, the function monotonously decreases or increases. Then the function needs to substitute the value of the extreme points of this segment. There are two ways.

1) Calculating the derivative and gaps on which it is positive or negative, to conclude that the function decreases on this segment or increases.

In accordance with them, substitute a larger or less argument value.

2) Just substitute both points into the function and compare the obtained function values.

In what tasks is the finding of the derivative optional

As a rule, in the tasks of the exam still need to find a derivative. There is only a couple of exceptions.

1) Parabola.

The top of Parabola is located according to the formula.

If A.< 0, то ветви параболы направлены вниз. И ее вершина является точкой максимума.

If a\u003e 0, then the parabola branches are directed upwards, the top is the minimum point.

Having calculated the point of the pearabela vertex, it is necessary to substitute its value to the function and calculate the corresponding function value.

2) Y \u003d TG X function. Or y \u003d CTG x.

These functions are monotonously increasing. Therefore, the greater the value of the argument, the greater the value of the function itself. Next, we will look at how to find the largest and smallest value of the function on the segment with examples.

Main types of tasks

Task: The greatest or smallest value of the function. Example on the chart.

In the figure you see a graph of the derivative function F (x) on the interval [-6; 6]. At what point is the segment [-3; 3] F (x) takes the smallest value?

So, to begin with, select the specified segment. On it, the function once takes a zero value and changes its mark - this is an extremum point. Since the derivative of the negative becomes positive, it means that the point is minimum function. This point corresponds to the value of the argument 2.

We continue to consider examples. Task: Find the greatest and smallest value of the function on the segment.

Find the smallest value of the function y \u003d (x - 8) E X-7 on the segment.

1. Take a derived from a complex function.

y "(x) \u003d (x - 8) e x-7 \u003d (x - 8)" (E X-7) + (x - 8) (E X-7) "\u003d 1 * (E X-7) + (x - 8) (E X-7) \u003d (1 + x - 8) (E x-7) \u003d (x - 7) (E X-7)

2. To equate the resulting derivative to zero and solve the equation.

(x - 7) (E X-7) \u003d 0

x - 7 \u003d 0, or e x-7 \u003d 0

x \u003d 7; E X-7 ≠ 0, no roots

3. Substitute the value of the extreme points, as well as the roots of the equation.

y (6) \u003d (6 - 8) E 6-7 \u003d -2E -1

y (7) \u003d (7 - 8) E 7-7 \u003d -1 * E 0 \u003d -1 * 1 \u003d -1

y (8) \u003d (8 - 8) E 8-7 \u003d 0 * E 1 \u003d 0

So, in this article, the main theory was considered on how to find the smallest value of the function on the segment required to successfully solve the tasks of the EGE on profile mathematics. Also, elements of mathematical analysis are used in solving tasks from the part from the exam, but obviously, they represent a different level of complexity, and their solutions algorithms are difficult to fit into the framework of one material.

Let's see how to explore the function using the graph. It turns out that looked at the schedule, you can find out everything that interests us, namely:

function definition area
function values \u200b\u200barea
zero function
gaps of increasing and descending
maximum and minimum points
the greatest and smallest value of the function on the segment.

Clarify terminology:

Abscissa - This is the horizontal point coordinate.
Ordinate - vertical coordinate.
Axis abscissa - horizontal axis, most often called the axis.
Axis ordinate - Vertical axis, or axis.

Argument - An independent variable on which the values \u200b\u200bof the function depend. Most often is indicated.
In other words, we ourselves choose, substitute the function in the formula and get.

Domain Functions are a set of those (and only those) of the values \u200b\u200bof the argument, in which the function exists.
Designated: or.

In our figure, the field definition area is a segment. It is on this segment that a function is drawn. Only here this function exists.

Function values \u200b\u200barea - This is a set of values \u200b\u200bthat take the variable. In our figure it is a segment - from the lowest to the highest value.

Zero function - Points where the value of the function is zero, that is. On our drawing is points and.

The values \u200b\u200bof the function are positive where . In our drawing, these are gaps and.
The values \u200b\u200bof the function are negative where . We have this gap (or interval) from to.

The most important concepts - ascending and decrease of function At some set. You can take a segment, interval, integration of the gaps or the entire numerical direct.

Function increases

In other words, the more, the more, that is, the schedule goes to the right and up.

Function decrease On the set, if for any and owned by the set, the inequality follows inequality.

For decreasing function, a greater value corresponds to a smaller value. The schedule goes to the right and down.

In our figure, the function increases on the interval and decreases at intervals and.

We define what maximum point and minimum function.

Maximum point - This is the inner point of the definition area, such that the value of the function in it is greater than in all the points close to it.
In other words, the maximum point is such a point, the value of the function in which morethan in the neighboring. This is a local "holmik" on the chart.

In our drawing - the point of the maximum.

Point of minimum - the inner point of the definition area, such that the value of the function in it is less than in all the points close to it.
That is, a minimum point is such that the value of the function in it is less than in the neighboring. On the schedule it is a local "fossa".

In our drawing - a minimum point.

Point is a boundary. It is not an internal point of the definition area and therefore does not suit the definition of a maximum point. After all, she has no neighbors on the left. Similarly, on our schedule there can be no point of minimum.

Maximum and minimum points are called points of extremum function. In our case, it is.

And what to do if you need to find, for example, minimum function On the segment? In this case, the answer :. Because minimum function - This is its value at a minimum point.

Similarly, the maximum of our function is equal. It is achieved at the point.

It can be said that the extremes of the function are equal and.

Sometimes in tasks you need to find the greatest and smallest values \u200b\u200bof the function On a given segment. They do not necessarily coincide with extremes.

In our case the smallest meaning of the function On the segment equals and coincides with the minimum function. But its greatest value on this segment is equal. It is achieved in the left end of the segment.

In any case, the largest and smallest values \u200b\u200bof the continuous function on the segment are achieved either at extremum points, or at the ends of the segment.

The standard algorithm for solving such tasks implies after finding the zeros of the function, determining the signs of the derivative on the intervals. Then the calculation of the values \u200b\u200bat the found points of the maximum (or a minimum) and on the interval boundary, depending on which the question is in the condition.

I advise you to do a little differently. Why? Wrote about it.

I propose to solve such tasks as follows:

1. Find a derivative.
2. Find zeros derivative.
3. Determine which of them belong to this interval.
4. Calculate the values \u200b\u200bof the function at the interval boundaries and points of claim 3.
5. We conclude (responding to the question).

During the solution of the examples presented, the solution of square equations is not considered in detail, it should be able to do. Also should know.

Consider examples:

77422. Locate the greatest value of the function y \u003d x 3 -3x + 4 on the segment [-2; 0].

We find zeros derivative:

The interval specified in the condition belongs to the point x \u003d -1.

Calculate the values \u200b\u200bof the function at points -2, -1 and 0:

The greatest value of the function is 6.

Answer: 6.

77425. Find the smallest value of the function y \u003d x 3 - 3x 2 + 2 on the segment.

Find a derivative of a given function:

We find zeros derivative:

The interval specified in the condition belongs to the point x \u003d 2.

Calculate the values \u200b\u200bof the function at points 1, 2 and 4:

The smallest function value is -2.

Answer: -2.

77426. Locate the greatest value of the function y \u003d x 3 - 6x 2 on the segment [-3; 3].

Find a derivative of a given function:

We find zeros derivative:

The interval specified in the condition belongs point x \u003d 0.

Calculate the values \u200b\u200bof the function at points -3, 0 and 3:

The smallest function value is 0.

Answer: 0.

77429. Find the smallest value of the function y \u003d x 3 - 2x 2 + x +3 on the segment.

Find a derivative of a given function:

3x 2 - 4x + 1 \u003d 0

We obtain the roots: x 1 \u003d 1 x 1 \u003d 1/3.

The interval specified in the condition belongs only x \u003d 1.

Find the values \u200b\u200bof the function at points 1 and 4:

Received that the smallest function value is 3.

Answer: 3.

77430. Find the highest value of the function y \u003d x 3 + 2x 2 + x + 3 on the segment [- 4; -one].

Find a derivative of a given function:

We will find the derivative of the derivative, solve the square equation:

3x 2 + 4x + 1 \u003d 0

We get roots:

The interval specified in the condition owns the root x \u003d -1.

We find the values \u200b\u200bof the function at points -4, -1, -1/3 and 1:

Received that the greatest value of the function is 3.

Answer: 3.

77433. Find the smallest value of the function y \u003d x 3 - x 2 - 40x +3 on the segment.

Find a derivative of a given function:

We will find the derivative of the derivative, solve the square equation:

3x 2 - 2x - 40 \u003d 0

We get roots:

The interval specified in the condition owns the root x \u003d 4.

We find the values \u200b\u200bof the function at points 0 and 4:

Received that the smallest function value is -109.

Answer: -109

Consider a method for determining the greatest and smallest values \u200b\u200bof the functions without a derivative. This approach can be used if you have big problems with the definition of the derivative. The principle is simple - to the function we substitute all the integers from the interval (the fact is that in all such prototypes the answer is an integer).

77437. Find the smallest value of the function y \u003d 7 + 12x-x 3 on the segment [-2; 2].

We substitute the points from -2 to 2: Watch the decision

77434. Locate the greatest value of the function y \u003d x 3 + 2x 2 - 4x + 4 on the segment [-2; 0].

That's all. Success to you!

Sincerely, Alexander Krutitsky.

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

Problem Statement 2:

Dana function, defined and continuous at some interval. It is required to find the largest (smallest) value of the function at this interval.

Theoretical basis.
Theorem (second Weierstrass theorem):

If the function is determined and continuous in a closed gap, then it reaches its largest and smallest values.

The function can reach its largest and smallest values \u200b\u200bor at the inner points of the gap, or on its borders. We will illustrate all possible options.

Explanation:
1) The function reaches its greatest value on the left border of the gap at the point, and its smallest value on the right border of the gap at the point.
2) The function reaches its highest value at the point (this is a maximum point), and its smallest value on the right border of the gap at the point.
3) The function reaches its highest value on the left border of the gap at the point, and its smallest value at the point (this is a minimum point).
4) The function is constant at the interval, i.e. It reaches its minimum and maximum value at any point of the gap, and the minimum and maximum values \u200b\u200bare equal to each other.
5) The function reaches its highest value at the point, and its smallest point value (despite the fact that the function has at this gap as maximum and at least).
6) The function reaches its highest value at the point (this is a maximum point), and its smallest value at the point (this is a minimum point).
Comment:

"Maximum" and "Maximum meaning" - different things. This follows from the determination of the maximum and intuitive understanding of the phrase "maximum meaning."

Algorithm for solving problems 2.

4) Choose from the most values \u200b\u200bof the largest (smallest) and write down the answer.

Example 4:

Determine the greatest and smallest function On the segment.
Decision:
1) Find a derived function.

2) Find stationary points (and points, suspicious to extremum), solving the equation. Pay attention to points in which there is no two-sided finite derivative.

3) Calculate the values \u200b\u200bof the function in stationary points and on the interval boundaries.

4) Choose from the most values \u200b\u200bof the largest (smallest) and write down the answer.

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

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

In the correctness of the calculations, you can make sure to look at the schedule of the function under study.

Comment: The greatest value reaches the maximum at the point, and the smallest is on the cut border.

Private case.

Suppose you need to find the most and minimum value of some function on the segment. After the first point of the algorithm, i.e. The calculation of the derivative becomes clear that, for example, it takes only negative values \u200b\u200bon the entire segment under consideration. Remember that if the derivative is negative, then the function decreases. Received that the function decreases on the whole segment. This situation is displayed in graph number 1 at the beginning of the article.

On the segment, the function decreases, i.e. She has no extremes. From the picture you can see that the smallest value of the function will take on the right border of the segment, and the greatest value is on the left. If the derivative on the segment is positive everywhere, then the function increases. The smallest meaning is on the left border of the segment, the greatest - on the right.

And to solve it will require minimal knowledge of the theme. Another academic year ends, everyone wants to break on vacation, and to bring this moment to bring this moment, I immediately turn to the case:

Let's start with the area. The area of \u200b\u200bwhich is spent in the condition is limited closed Many points of the plane. For example, a set of points limited by a triangle, including the whole triangle (if because borders "To buy" at least one point, the region will cease to be closed). In practice, there are also areas of rectangular, round and slightly more complex forms. It should be noted that there are strict definitions in the theory of mathematical analysis. limitations, closers, borders, etc.But I think everyone is aware of these concepts at an intuitive level, and more and now do not.

The flat area is standardly denoted by the letter, and, as a rule, is set analytically - several equations (not necessarily linear); less often inequalities. Typical verbal turnover: "Closed area limited by lines."

An integral part of the task under consideration is to build the area in the drawing. How to do it? You need to draw all of the listed lines (in this case 3 straight) And analyze what happened. The desired area is usually slightly stroking, and its border is distinguished by a bold line:

The same area can be set and linear inequalities: that for some reason more often write down by the transition list, and not system.
Since the border belongs to the region, then all inequalities, of course, neztreat.

And now the essence of the task. Imagine that from the beginning of the coordinates, the axis goes directly. Consider a function that continuous in each Point of region. The schedule of this function is some surfaceAnd little happiness is that to solve today's task, we do not need to know at all how this surface looks like. It can be placed above, below, cross the plane - all this is not important. And the following is: according to weierstrass theorems, continuous in limited closedareas The function reaches the greatest ("High") and the smallest ("Low" himself) The values \u200b\u200bthat are required to find. Such values \u200b\u200bare achieved or in stationary points, owned areasD. , orat points that lie on the border of this area. What follows a simple and transparent solution algorithm:

Example 1.

In a limited closed area

Decision: First of all, you need to portray the area in the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the task, and therefore I will immediately give the final illustration, which shows all the "suspicious" points found during the study. Usually they are affixed one by the other as they are detected:

Based on the preamble, the solution is convenient to smash two points:

I) find stationary points. This is the standard action that we have repeatedly performed at the lesson. about extremes of several variables:

Found Stationary Point belongs Areas: (We celebrate it in the drawing)So, we should calculate the value of the function at this point:

- as in the article The greatest and smallest values \u200b\u200bof the function on the segmentImportant results I will highlight bold font. In the notebook they are convenient to circle a pencil.

Pay attention to our second happiness - there is no point in checking a sufficient condition of Extremum. Why? Even if the function reaches a function, for example, local minimumthen it does not mean that the obtained value will be minimal In the whole area (see the beginning of the lesson on unconditional extremumums) .

What if the stationary point does not belong to the region? Almost nothing! It should be noted that and go to the next item.

(Ii) Explore the border of the region.

Since the border consists of the sides of the triangle, then the study is convenient to split into 3 subparagraphs. But it is better to do it not ababa like. From my point of view, first it is more advantageous to consider the segments parallel to the coordinate axes, and first of all - lying on the axes themselves. To catch the entire sequence and logic of actions Try to learn the ending "in one breath":

1) We will deal with the bottom side of the triangle. To do this, we will substitute directly to the function:

Alternatively, you can arrange and so:

Geometrically, this means that the coordinate plane (which is also set by the equation) "Carves" from surface "Spatial" parabola, whose vertex immediately falls under suspicion. Find out where she is located:

- the resulting value "hit" to the area, and may well be that at the point (celebrate in the drawing) The function reaches the greatest or smallest value in the entire area. Anyway, carry out computation:

Other "candidates" are, of course, the ends of the segment. Calculate the values \u200b\u200bof the function at points (celebrate in the drawing):

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

2) For the study of the right side of the triangle, we substitute the function and "order there":

Here immediately perform a draft check, "nicknamed" the end of the segment is already treated:
, well.

The geometric situation is related to the previous item:

- The resulting value also "went into the sphere of our interests", which means it is necessary to calculate what is equal to the function in the point that appears:

We explore the second end of the segment:

Using a function , Perform a check check:

3) Probably everyone is guessing how to explore the rest. We substitute the function and make simplifications:

Ends of cut already investigated, but on the draft still check, whether we have 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 the line to the equation, we get the ordinate of this "Interest":

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

Check the calculations on the "budget" version :
, order.

And the final step: Carefully view all the "fat" numbers, which begin to recommend even to compile a single list:

From which we choose the greatest and smallest meanings. Answer We write in the style of the task of staying the greatest and smallest values \u200b\u200bof the function on the segment:

Just in case, once again comment on the geometric meaning of the result:
- Here is the highest surface point in the area;
- Here is the lowest point of the surface in the area.

In the disassembled task, we have already revealed 7 "suspicious" points, but from the task of the task, their number varies. For the triangular area, the minimum "research set" consists of three points. This happens when the function, for example, asks plane - It is absolutely clear that stationary points are absent, and the function can achieve the greatest / smallest values \u200b\u200bonly in the vertices of the triangle. But such examples once, two and turned around - usually have to deal with some the surface of the 2nd order.

If you make such assignments a little, then from the triangles the head can go around, and therefore I prepared for you unusual examples so that it becomes square :))

Example 2.

Find the greatest and smallest values \u200b\u200bof the function in a closed area limited lines

Example 3.

Find the greatest and smallest values \u200b\u200bof the function in a limited closed area.

Pay special attention to the rational order and technique of studying the boundaries of the region, as well as on the chain of intermediate checks, which will almost absolutely allow to avoid computing errors. Generally speaking, you can solve as you like, but in some tasks, for example, in the same example 2, there are all chances to significantly complicate your life. An exemplary sample of finishing tasks at the end of the lesson.

We systematize the algorithm of the solution, and then with my diligence of spider, he somehow got lost in the long thread comments of the 1st Example:

- In the first step we build an area, it is desirable to shake it, and the boundary is to highlight the bold line. During the solution, points will appear to be installed in the drawing.

- find stationary points and calculate the values \u200b\u200bof the function only in those of themwhich belong to the area. The obtained values \u200b\u200bare separated in the text (for example, supply a pencil). If the stationary point does not belong to the region, then we celebrate this fact a badge or verbally. If there are no stationary points at all, then we make a written conclusion that they are missing. In any case, this item cannot be skipped!

- Explore the border of the region. First, it is advantageous to deal with straight, which are parallel to the coordinate axes (if there are any). The values \u200b\u200bof the function calculated in the "suspicious" points also allocate. About the technique of solutions is very much told above and something else will be said below - read, re-read, delve it!

- From the selected numbers, choose the largest and smallest values \u200b\u200band give the answer. Sometimes it happens that such values \u200b\u200bfeature reaches at once at several points - in this case all these points should be reflected in the response. Let, for example, And it turned out that this is the smallest meaning. Then write down that

Final examples are devoted to other useful ideas that will be useful in practice:

Example 4.

Find the greatest and smallest values \u200b\u200bof the function in a closed area .

I retained the copyright formulation in which the region is asked in the form of double inequality. This condition can be recorded by an equivalent system or in a more traditional form for this task:

I remind you that with nonlinear Inequalities we encountered on, and if you do not understand the geometric meaning of the record, then please do not postpone and clarify the situation right now ;-)

DecisionAs always, begins with the construction of the region, which is a kind of "sole":

Hmm, sometimes you have to nibble not only granite science ....

I) find stationary points:

Idiot's dream system :)

The stationary point belongs to the region, namely, lies at its border.

And so, it, nothing ... The lesson went to go - that's what it means to drink the right tea \u003d)

(Ii) Explore the border of the region. Without the caustav, we start with the abscissa axis:

1) if, then

We will find where the top of Parabola:
- Appreciate such moments - "got" directly to the point with which everything is already clear. But you still do not forget about checking:

We calculate the values \u200b\u200bof the function at the ends of the segment:

2) with the bottom of the "soles" will understand "for one sitting" - without any complexes we substitute to function, and we will only be interested in the segment:

Control:

This is already contributing some revival into a monotonous ride at the rolled rut. Find critical points:

Decide quadratic equation, remember about such? ... However, remember, of course, otherwise they would not read these lines \u003d) if in the two previous examples there were convenient calculations in decimal fractions (which, by the way, rarity), here we will be waiting for the usual ordinary fractions. We find "Icx" roots and by the equation, we define the corresponding "ignorable" coordinates of the points "candidates":

Calculate the functions of the function at found points:

Specify the function yourself.

Now carefully study the conquered trophies and write down answer:

These are "candidates", so "candidates"!

For self solutions:

Example 5.

Find the smallest and greatest values \u200b\u200bof the function in a closed area

Recording with figured brackets is read like this: "Many points, such as".

Sometimes in such examples use lagrange multiplier methodBut the real need to apply it is unlikely to arise. For example, if a function is given with the same area "DE", then after the substitution in it - with a derivative of any difficulties; And it is drawn up with a "one line" (with signs) without need to consider the upper and lower semicircle separately. But, of course, there are more difficult cases where without the Lagrange function (where, for example, the same circumference equation) It is difficult to do without it - how difficult it is to do without a good rest!

Everyone is well to pass the session to the soon meetings next season!

Solutions and answers:

Example 2: Decision: Show area in the drawing: