The study of such an object of mathematical analysis as a function has a large value and in other fields of science. For example, in economic analysis, you constantly need to evaluate behavior function profit, namely to determine its greatest value and develop a strategy to achieve it.
Instructions
Anyone's behavior research should always begin with finding a domain. Usually, according to the condition of a specific task, it is required to determine the greatest value function either over this entire area, or over a specific interval with open or closed boundaries.
Based on, the largest is value function y (x0), for which the inequality y (x0) ≥ y (x) (х ≠ x0) holds for any point of the domain of definition. Graphically, this point will be the highest if you arrange the values \u200b\u200bof the argument along the abscissa, and the function itself along the ordinate.
To determine the greatest value function, follow the three-step algorithm. Note that you must be able to work with one-way and, and also calculate the derivative. So, let some function y (x) be given and it is required to find its greatest value on some interval with boundary values \u200b\u200bA and B.
Find out if this interval is within the scope function... To do this, you need to find it, having considered all possible restrictions: the presence in the expression of a fraction, square root, etc. Scope is the set of argument values \u200b\u200bfor which a function makes sense. Determine if the given interval is a subset of it. If so, go to the next step.
Find the derivative function and solve the resulting equation by equating the derivative to zero. Thus, you get the values \u200b\u200bof the so-called stationary points. Estimate whether at least one of them belongs to the interval A, B.
Consider these points at the third stage, substitute their values \u200b\u200binto the function. Perform the following additional steps depending on the type of interval. If there is a segment of the form [A, B], the boundary points are included in the interval, this is indicated by the brackets. Calculate the values function at x \u003d A and x \u003d B. If the open interval is (A, B), the boundary values \u200b\u200bare punctured, i.e. are not included in it. Solve the one-sided limits for x → A and x → B. A combined interval of the form [A, B) or (A, B], one of the boundaries of which 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, act according to the principles already described, and for infinite look for limits for x → -∞ and x → + ∞, respectively.
The task at this stage
And to solve it, you need minimal knowledge of the topic. The next school year is coming to an end, 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 of the plane. For example, a set of points bounded by a triangle, including the WHOLE triangle (if from boundaries "Gouge" at least one point, then the area will cease to be closed)... In practice, there are also areas of rectangular, round and slightly more complex shapes. It should be noted that strict definitions are given in the theory of mathematical analysis limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and more is not needed now.
A flat area is denoted by a letter in the standard way, and, as a rule, it is set analytically - by several equations (not necessarily linear); less often inequalities. Typical turnover: "closed area, bounded by lines."
An integral part of the task under consideration is the construction of an area in 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 slightly hatched, and its border is highlighted with a bold line: 
The same area can be set and linear inequalities:, which for some reason are more often written as an enumerated list, and not system.
Since the boundary belongs to the region, all inequalities, of course, lax.
And now the essence of the problem. Imagine an axis extending from the origin directly towards you. Consider a function that continuous in each point of the area. The graph of this function represents some surface, and a small happiness lies in the fact that to solve today's problem, we do not need to know what this surface looks like. It can be located higher, lower, intersect the plane - all this is not important. And the following is important: according to weierstrass theorems, continuous in limited closedarea, the function reaches the maximum (the "highest") and the smallest (the "lowest") values \u200b\u200bthat you want to find. Such values \u200b\u200bare achieved or in stationary points, belonging to the regionD , orat points that lie on the border of this area. From what follows a simple and transparent solution algorithm:
Example 1
In a limited enclosed area
Decision: First of all, you need to depict the area in 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 research. Usually they are affixed one after the other as they are found: 
Based on the preamble, it is convenient to split the decision into two points:
I) Find stationary points. This is a standard action that we have repeatedly performed in the lesson. 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 this point:
- as in the article The largest and smallest values \u200b\u200bof the function on the segment, I will highlight the important results in bold. In a notebook it is convenient to outline them with a pencil.
Pay attention to our second happiness - there is no point in checking sufficient condition for extremum... Why? Even if at a point the function reaches, for example, local minimum, then it STILL DOES NOT MEAN that the resulting value will be minimal throughout the region (see the beginning of the lesson about unconditional extrema) .
What if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and go to the next item.
II) Explore the boundary of the region.
Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 sub-items. But it's better not to do it 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 - lying on the axes themselves. To catch the whole sequence and logic of actions, try to study the ending "in one go":
1) Let's deal with the bottom side of the triangle. To do this, we substitute directly into the function:
Alternatively, you can arrange it like this: 
Geometrically, this means that the coordinate plane (which is also given by the equation) "Carves" out surfaces A "spatial" parabola, the vertex of which immediately comes under suspicion. Let's find out where is she:
- the obtained value "got" into the area, and it may well be that at the point (mark in the drawing) the function reaches the highest or the lowest value in the entire area. One way or another, we carry out calculations:
Other "candidates" are, of course, the ends of the segment. Let's calculate the values \u200b\u200bof the function at points 
(mark in the drawing):
Here, by the way, you can perform a verbal mini-check using 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 will immediately perform a rough check, "ringing" the already processed end of the segment:
, well.
The geometric situation is related to the previous point:
- the resulting value is also "included in the scope of our interests", which means that we need to calculate what the function is equal to at the point that appears:
Let's examine the second end of the segment:
Using the function 
, let's check it out: 
3) Probably everyone knows how to explore the remaining side. We substitute in the function and perform simplifications:
Segment ends 
have already been investigated, but on the draft we still check if 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 in the drawing and find the corresponding value of the function:
Let's check the calculations according to the "budget" version 
:
, order.
And the final step: CAREFULLY we look through all the "fat" numbers, I recommend that beginners even make a single list: 
from which we choose the largest and smallest values. Answer we write in the style of the problem of finding the largest and smallest values \u200b\u200bof the function on the segment:
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 area;
Is the lowest surface point in the area.
In the analyzed problem, we identified 7 "suspicious" points, but their number varies from task to task. For a triangular region, the minimum "research set" is three points. This happens when a function, for example, sets plane - it is quite clear that there are no stationary points, and the function can reach the largest / smallest values \u200b\u200bonly at the vertices of the triangle. But there are a lot of such examples once or twice - usually you have to deal with some surface of the 2nd order.
If you solve such tasks a little, then triangles may make your head spin, and therefore I have prepared unusual examples for you to make it square :))
Example 2
Find the largest and smallest function values 
in a closed area bounded by lines
Example 3
Find the largest and smallest values \u200b\u200bof a function in a bounded closed area.
Pay special attention to the rational order and technique for examining the region boundary, 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.
Let's systematize the solution algorithm, otherwise, with my diligence as a spider, it somehow got lost in the long thread of comments from 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 placed on the drawing.
- Find stationary points and calculate the values \u200b\u200bof the function only in those of themthat belong to the area. Highlight the resulting values \u200b\u200bin the text (for example, outline them with a pencil). If the stationary point does NOT belong to the region, 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!
- Let's explore the border of the area. First, it is beneficial to deal with straight lines that are parallel to the coordinate axes (if any)... We also highlight the values \u200b\u200bof the function calculated at "suspicious" points. A lot has been said above about the solution technique 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 the answer. Sometimes it happens that the function reaches such values \u200b\u200bat several points at once - in this case, all these points should be reflected in the answer. Let, for example, 
and it turned out to be the smallest value. Then we write down 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 \u200b\u200bof a function in a closed area 
.
I have kept the author's formulation, in which the region is given as a double inequality. This condition can be written by an equivalent system or in a more traditional form for this problem: 
I remind you that since nonlinear inequalities we encountered on, and if you do not understand the geometric meaning of the notation, then please do not postpone and clarify the situation right now ;-)
Decision, as always, it starts with building an area, which is a kind of "sole": 
Hmm, sometimes you have to gnaw not only the granite of science….
I) Find stationary points: 
System-idiot's dream :) 
A stationary point belongs to the region, namely, lies on its boundary.
And so, it is, nothing ... the lesson went cheerfully - that's what it means to drink the right tea \u003d)
II) Explore the boundary of the region. Without further ado, let's start with the abscissa:
1) If, then
Let's find where the vertex of the parabola is:
- appreciate such moments - "hit" right at the point from which everything is already clear. But don't forget about checking: 
Let's calculate the values \u200b\u200bof 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:
This already brings some revival to the monotonous driving on the knurled track. Let's find the critical points:
We solve quadratic equation, remember this one more? ... However, remember, of course, otherwise you would not have read these lines \u003d) If in the two previous examples calculations in decimal fractions were convenient (which, by the way, is a rarity), here we are waiting for the usual ordinary fractions. We find the “x” roots and use the equation to determine the corresponding “game” coordinates of the “candidate” points: 
Let's calculate the values \u200b\u200bof the function at the found points: 
Check the function yourself.
Now we carefully study the won trophies and write down answer:
These are "candidates", so "candidates"!
For an independent solution:
Example 5
Find the smallest and largest values \u200b\u200bof a function 
in a closed area 
An entry with curly braces reads like this: "many points, such that".
Sometimes in such examples they use lagrange multiplier method, but the real need to apply it is unlikely to arise. So, for example, if a function is given with the same domain "de", then after substitution into it - with a derivative of no difficulties; moreover, everything is drawn up "in one line" (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are also more complex cases where, without the Lagrange function (where, for example, the same equation of the circle) it is difficult to manage - how difficult it is to do without a good rest!
It's good for everyone to pass the session and see you soon next season!
Solutions and Answers:
Example 2: Decision: depict the area in the drawing:
Page 1
Theoretical facts:
The square trinomial \u003d ax2 + bx + c has an extreme value taken by it at
This value turns out to be the smallest if a\u003e 0, and the largest if a< 0. Если существует y(макс), то y(мин) не существует, и наоборот.
# 1. Decompose a given positive number A into two terms so that their product turns out to be the largest.
Decision. Let us denote one of the required terms by x. Then the second term will be equal to A - x, and their product 
or. 
Thus, the question led to finding such a value of x at which this square trinomial will receive the greatest value. According to Theorem 4, such a value certainly exists (because here the leading coefficient is equal to - 1, i.e., negative) and is equal to In this case, and, therefore, both terms must be equal to each other.
For example, the number 30 allows for the following expansions:
All received works are less than
# 2. There is a wire of length L. It is required to bend it so that a rectangle is obtained, limiting the largest possible area.
Decision. Let us denote (Fig. 1) one of the sides of the rectangle through x. Then, obviously, its other side will be a 
or 
... This function takes on its greatest value at, which will be the desired value of one of the sides of the rectangle. Then its other side will be, that is, our rectangle turns out to be a square. The obtained solution to the problem can be summarized in the form of the following theorem.
Of all the rectangles that have the same perimeter, the square has the largest area.
Comment.
Our problem is also easy to solve using the result obtained in solving problem 1.
Indeed, we see that the area of \u200b\u200bthe rectangle of interest to us is In other words, there is a product of two factors x and But the sum of these factors is 
, t. that is, a number that does not depend on the choice of x. This means that the matter is reduced to the decomposition of a number into two terms so that their product is the largest. As we know, this product will be the largest when both terms are equal, i.e.
No. 3. From the available boards, you can build a fence 200 m long. It is required to enclose a rectangular yard of the largest area with this fence, using a factory wall for one side of the yard.
trinomial theorem derivative function
Decision. Let us denote (Fig. 2) one of the sides of the yard through x. Then its other side will be equal and its area will be
According to the theorem, the largest value of this function is attained by it for
So, the side of the yard, perpendicular to the factory wall, should be 50 m, whence for the side parallel to the wall, the value is 100 m, that is, the yard should have the shape of half a square.
From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest values \u200b\u200bof a function. What is the reason for this? Profit maximization, cost minimization, determination of the optimal equipment load ... In other words, in many spheres of life one has to solve the problem of optimizing any parameters. And these are the tasks of finding the largest and smallest values \u200b\u200bof a function.
It should be noted that the largest and smallest value of a function is usually searched for in some interval X, which is either the entire domain of the function or part of the domain. The X interval itself can be a segment, an open interval 
, an endless interval.
In this article we will talk about finding the largest and smallest values \u200b\u200bof an explicitly given function of one variable y \u003d f (x).
Page navigation.
The highest and the lowest value of the function - definitions, illustrations.
Let us briefly dwell on the main definitions.
The largest value of the function 
that for any 
the inequality is true.
Smallest function value y \u003d f (x) on the interval X is called such a value 
that for any the inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) accepted value on the considered interval at the abscissa.
Stationary points Are the argument values \u200b\u200bat which the derivative of the function vanishes.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its largest (smallest) value on the interval X at one of the stationary points from this interval.
Also, often, a function can take the largest and smallest value at points where the first derivative of this function does not exist, and the function itself is defined.
Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of definition of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the highest and the lowest value of the function.
For clarity, we will give a graphic illustration. Look at the pictures and a lot will become clear.
On the segment
In the first figure, the function takes the largest (max y) and the smallest (min y) values \u200b\u200bat stationary points inside the segment [-6; 6].
Consider the case shown in the second figure. Change the segment to. In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.
In Figure 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.
On an open interval
In the fourth figure, the function takes the largest (max y) and smallest (min y) values \u200b\u200bat stationary points located within the open interval (-6; 6).
On the interval, no conclusions can be drawn about the largest value.
At infinity
In the example shown in the seventh figure, the function takes the largest value (max y) at a stationary point with the abscissa x \u003d 1, and the smallest value (min y) is reached at the right border of the interval. At minus infinity, the values \u200b\u200bof the function asymptotically approach y \u003d 3.
In the interval, the function does not reach either the smallest or largest value. When tending to x \u003d 2 on the right, the values \u200b\u200bof the function tend to minus infinity (the straight line x \u003d 2 is the vertical asymptote), and when the abscissa tends to plus infinity, the values \u200b\u200bof the function asymptotically approach y \u003d 3. A graphic illustration of this example is shown in Figure 8.
Algorithm for finding the largest and smallest values \u200b\u200bof a continuous function on a segment.
Let's write an algorithm that allows us to find the largest and smallest value of a function on a segment.
- Find the domain of the function and check if it contains the entire segment.
- We find all the points at which the first derivative does not exist and which are contained in the segment (usually such points are found in functions with an argument under the modulus sign and in power functions with fractional rational exponent). If there are no such points, then go to the next item.
- Determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them falls into the segment, then go to the next item.
- We calculate the values \u200b\u200bof the function at the selected stationary points (if any), at the points where the first derivative does not exist (if any), as well as for x \u003d a and x \u003d b.
- From the obtained values \u200b\u200bof the function, we select the largest and the smallest - they will be the desired largest and smallest values \u200b\u200bof the function, respectively.
Let us analyze the algorithm when solving an example for finding the largest and smallest values \u200b\u200bof a function on a segment.
Example.
Find the largest and smallest function value
- on the segment;
- on the segment [-4; -1].
Decision.
The domain of a function is the entire set of real numbers, except for zero, that is. Both segments fall within the definition area.
Find the derivative of the function with respect to: 
Obviously, the derivative of the function exists at all points of the segments and [-4; -1].
The stationary points are determined from the equation. The only valid root is x \u003d 2. This stationary point falls into the first segment.
For the first case, we calculate the values \u200b\u200bof the function at the ends of the segment and at the stationary point, that is, for x \u003d 1, x \u003d 2 and x \u003d 4: 
Therefore, the largest value of the function 
is attained at x \u003d 1, and the smallest value 
- for x \u003d 2.
For the second case, we calculate the values \u200b\u200bof the function only at the ends of the segment [-4; -1] (since it does not contain a single stationary point): 
Sometimes problems B15 come across "bad" functions for which it is difficult to find a derivative. Previously, this was only on probes, but now these tasks are so common that they can no longer be ignored when preparing for the real exam.
In this case, other tricks work, one of which is - monotone.
A function f (x) is called monotonically increasing on a 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 (x 2).
A function f (x) is called monotonically decreasing on a segment if the following holds for any points x 1 and x 2 of this segment:
x 1< x 2 ⇒ f (x 1)\u003e f ( x 2).
In other words, for an increasing function, the larger the x, the larger the f (x). For a decreasing function, the opposite is true: the larger x, the less f (x).
For example, the logarithm increases monotonically if the base a\u003e 1, and decreases monotonically if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.
f (x) \u003d log a x (a\u003e 0; a ≠ 1; x\u003e 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 grows for a\u003e 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) \u003d a x (a\u003e 0)
Finally, negative exponents. You can write them as a fraction. Have a discontinuity point at which monotony is broken.
All these functions are never found in their pure form. They add polynomials, fractions and other nonsense, because of which it becomes difficult to count 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 \u003d ax 2 + bx + c. Its graph is a standard parabola in which we are interested in:
- Parabola branches - can go up (for a\u003e 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
- The vertex of a parabola is the extremum point of a quadratic function at which this function takes its smallest (for a\u003e 0) or largest (a< 0) значение.
Of greatest interest is precisely apex 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 will formulate the key rule:
The extremum points of the quadratic trinomial and the complex function it enters into coincide. Therefore, you can search for x 0 for a square trinomial, and score on a function.
From the above reasoning, it remains unclear which point we get: a maximum or a minimum. However, the tasks are specially designed so that it does not matter. Judge for yourself:
- There is no segment in the problem statement. Therefore, you don't need to compute f (a) and f (b). It remains to consider only the extremum points;
- But there is only one such point - this is the vertex of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.
Thus, the solution to the problem is greatly simplified and comes down to just two steps:
- Write out the equation of the parabola y \u003d ax 2 + bx + c and find its vertex by the formula: x 0 \u003d −b / 2a;
- 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 rationale may seem complex. I deliberately do not post a "bare" solution scheme, since thoughtless application of such rules is fraught with errors.
Consider the real problems from the trial exam in mathematics - this is where this technique is most often encountered. At the same time, we will make sure that in this way many B15 tasks become almost verbal.
Under the root is the 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.
The 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 the smallest value.
The root increases monotonically, so x 0 is the minimum point of the entire function. We have:
A task. Find the smallest value of the function:
y \u003d log 2 (x 2 + 2x + 9)
Under the logarithm, there is again a quadratic function: y \u003d x 2 + 2x + 9. The graph is a parabola with branches up, since a \u003d 1\u003e 0.
The top of the parabola:
x 0 \u003d −b / (2a) \u003d −2 / (2 1) \u003d −2/2 \u003d −1
So, at the point x 0 \u003d −1 the quadratic function takes the smallest value. But the function y \u003d log 2 x is monotonic, therefore:
y min \u003d y (−1) \u003d log 2 ((−1) 2 + 2 (−1) + 9) \u003d ... \u003d log 2 8 \u003d 3
The exponent contains the quadratic function y \u003d 1 - 4x - x 2. Rewrite it in its normal form: y \u003d −x 2 - 4x + 1.
Obviously, the graph of this function is a parabola, branches down (a \u003d −1< 0). Поэтому вершина будет точкой максимума:
x 0 \u003d −b / (2a) \u003d - (- 4) / (2 (−1)) \u003d 4 / (- 2) \u003d −2
The original function is exponential, it is monotonic, so the largest value will be at the found point x 0 \u003d −2:
The attentive reader will probably notice that we did not write out the range of valid values \u200b\u200bof the root and logarithm. But this was not required: inside there are functions whose values \u200b\u200bare always positive.
Consequences from the domain of the function
Sometimes finding the vertex of the parabola is not enough to solve Problem B15. The desired value may lie at the end of the segment, but not at the extremum point. If there is no segment specified in the problem at all, look at range of valid values the original function. Namely:
Note again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how this works with specific examples:
A 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 downward, since a \u003d −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический квадратный корень из отрицательного числа не существует.
We write out the range of permissible values \u200b\u200b(ODZ):
3 - 2x - x 2 ≥ 0 ⇒ x 2 + 2x - 3 ≤ 0 ⇒ (x + 3) (x - 1) ≤ 0 ⇒ x ∈ [−3; one]
Now let's find the vertex of the parabola:
x 0 \u003d −b / (2a) \u003d - (- 2) / (2 (−1)) \u003d 2 / (- 2) \u003d −1
The point x 0 \u003d −1 belongs to the segment of ODZ - and this is good. Now we calculate the value of the function at the point x 0, as well as at the ends of the ODZ:
y (−3) \u003d y (1) \u003d 0
So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.
A task. Find the smallest value of the function:
y \u003d 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 can be no negative numbers in the logarithm, so we write the ODZ:
6x - x 2 - 5\u003e 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. This is how the logarithm differs from the root, where the ends of the segment are quite suitable for us.
We are looking for the top of the parabola:
x 0 \u003d −b / (2a) \u003d −6 / (2 (−1)) \u003d −6 / (- 2) \u003d 3
The vertex of the parabola is suitable for the ODV: x 0 \u003d 3 ∈ (1; 5). But since we are not interested in the ends of the segment, we consider the value of the function only at the point x 0:
y min \u003d y (3) \u003d log 0.5 (6 3 - 3 2 - 5) \u003d log 0.5 (18 - 9 - 5) \u003d log 0.5 4 \u003d −2