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Indeed, it is often necessary to beat the odds in order to calculate the most and least value of the function. We vikonuemo tse diyu todi, if it’s necessary, how to minimize vitrati, increase profits, optimize the optimal development of virobnitstva and others. To do so correctly, it is necessary to have a good understanding of what is the most and least important function.
Sound mi vyznaєmo tsі value at the borders of the deyago і interval x, which can with its own line show all the areas of the function of the yogo part. Tse mozhe buti yak vіdrіzok [a; b ] , i specified interval (a ; b) , (a ; b ) , [ a ; b) , infinite interval (a ; b) , (a ; b ) , [ a ; b) or indefinite interval - ∞ ; a , (- ∞ ; a ) , [ a ; + ∞) , (- ∞ ; + ∞) .
For each material, it is possible, how to calculate the most and least values of the explicitly given function with one variable y=f(x) y = f(x) .
Main appointments
Let's do it, as a rule, from the formulary of the main appointments. Appointment 1 The largest value of the function y = f (x) on the current interval x is the value m a x y = f (x 0) x ∈ X (x0). Appointment 2 The smallest value of the function y = f (x) on the current interval x is the value of m i n x ∈ X y = f (x 0) , so for any value x ∈ X , x ≠ x 0 ) ≥ f(x0) .
Qi vyznachennya є dosit obvious. More simply, you can say this: the largest value of the function is the largest value on the given interval at abscissa x 0, and the least - the least value is taken on the same interval at x 0 . Appointment 3 Stationary points are called such values of the argument of the function, for which it is likely to go up to 0.
We need to know what are stationary points? For the correct circuit, you need to guess Fermat's theorem. It is obvious that a stationary point is such a point, in which there is an extremum of a function that can be differentiated (that is a local minimum or a maximum). Also, the function is the least or the most significant on the singing interval itself in one of the stationary points.
Another function can be most or least significant at quiet points, for which the function itself is singing, but it is not the first one.
First of all, if you blame them on the following points: what can we assign the most or the least value of a function to a given score in all modes? Hі, we can not do it even if between the given interval the spacing is between the boundaries of the designated area, otherwise we can do it right with an indefinite interval. And so, that the function in a given context, or on infinity, takes infinitely small or infinitely large values. In these situations, it is impossible to assign the most and/or the least value.
The most mind-blowing moments become after the image on the graphs:
The first little one shows us the function, how to get the highest and the lowest value (m a x y і m i n y) at stationary points, spread on the rail [ - 6 ; 6].
Reportingly, we will analyze the types, appointments for another schedule. We change the value of the argument on [1; 6] and it is important that the greatest value of the function can be reached at the point with the abscissa at the right interinterval, and the least at the stationary point.
On the third little abscissa, the point is the boundary points of the vіdrіzka [-3; 2]. The stinks give the highest and the lowest value of the given function.
Now let's marvel at the fourth baby. For a new function, it takes m a x y (the largest value) and m i n y (the least value) at stationary points on a wide interval (-6; 6).
How do we take the interval [1; 6) , we can say that the smallest value of a function for a new one will be achieved at a stationary point. We will not know the greatest value. The function could take on the most value at x, which would be 6, but x = 6 would lie within the interval. The very peak is marked on the chart 5 .
On graph 6, the least value is given to the function of the right inter-interval (- 3 ; 2 ), and about the highest value, we cannot add the same vysnovkіv.
On the little one 7 Bachimo, that the function is matime m a x y at the stationary point, that the abscissa is equal to 1. The smallest value of the function is within reach of the interval on the right side. At minus inconsistency, the values of the function asymptotically approach up to y = 3 .
How can we take the interval x ∈ 2; + ∞ , then it is possible that the given function is not accepted for the newest or smallest, or largest value. If x is correct 2, then the value of the function is pragmatic minus the inconsistency, the scaling of the line x = 2 is the vertical asymptote. Although the abscissa is right up to plus inconsistency, then the value of the function asymptotically approximates up to y = 3 . The male of the vipadok is depicted as a baby 8 .
At this point, we will introduce a sequence of diy, as it is necessary to mark the highest and lowest value of the function on the singing voice.
- We know the scope of the assigned function. Pereverimo, chi to enter before her tasks for the mind of the wreckers.
- Now we can count the points that can be found in this wind, in some first place. Most often, it is possible to use functions, the argument of which entries is under the sign of the module, but for state functions, the indicator of which is a fractional rational number.
- Dali z'yasuєmo, yakі stationary points to spend at the tasks of vіdrіzok. For which you need to calculate the rest of the function, then equate it to 0 and the difference is equal, which happened in the result, after which you choose the appropriate root. Since we don’t see any stationary point, otherwise we won’t get the stink from the tasks of the breeches, we move on to the offensive croc.
- Significantly, if the value of the function is accepted at given stationary points (like stink є), or at quiet points, in which it is not the first time (like stink є), or the value for x = a і x = b.
- 5. We have a number of function values, from which it is now necessary to choose the most and least. What will be the most and least important functions that we need to know.
We wonder how correctly the algorithm is loaded for the first time of the day. butt 1 Umov: the function y = x3+4x2 is given. The most important and the least significant on the vіdrіzkah [1; 4] i [-4; -1].
Solution:
Let's look at the significance of the area assigned to this function. And here i will be impersonal of all real numbers, crim 0 . In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; +∞. Offenses, given in the mind, will be found in the middle of the designated area.
Now we can calculate the following functions according to the rule of fraction differentiation:
y "= x 3 + 4 x 2" = x 3 + 4 " x 2 - x 3 + 4 x 2" x 4 = = 3 x 2 x 2 - (x 3 - 4) 2 x x 4 = x 3 - 8 x 3
We found out that similar functions are available at all points in the openings [1; 4] i [-4; -1].
Now we need to determine the stationary points of the function. Zrobimo tse for additional help x 3 - 8 x 3 \u003d 0. The new one has only one real root, which is dear 2. Vіn will be a stationary point of function and eat at the first vіdrіzok [1; 4].
Let's calculate the value of the function of the first point and at the other point, tobto. for x = 1, x = 2 and x = 4:
y(1) = 1 3 + 4 1 2 = 5 y(2) = 2 3 + 4 2 2 = 3 y(4) = 4 3 + 4 4 2 = 4 1 4
We took away the largest value of the function m a x y x ∈ [1; 4 ] = y (2) = 3 will be reached in x = 1 , and at least m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – for x = 2 .
The other branch does not include any stationary point, so we need to calculate the values of the function only at the ends of the given branch:
y(-1) = (-1) 3 + 4 (-1) 2 = 3
So, m a x y x ∈ [- 4; - 1] = y (- 1) = 3, m i n y x ∈ [- 4; - 1] = y(-4) = - 3 3 4 .
Suggestion: For vіdrіzka [1; 4] - m a x y x ∈ [1; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 for the reverse [ - 4 ; - 1 ] - m a x y x ∈ [ - 4; - 1] = y (- 1) = 3, m i n y x ∈ [- 4; - 1] = y(-4) = - 3 3 4 .
for a little one:
Before that, how to learn the way, for the sake of repeating to you, how to correctly calculate the one-sided between and between on the inconsistency, as well as learn about the main methods of their recognition. In order to know the most and/or the least value of a function on a given or indefinite interval, it is necessary to sequentially do so.
- For the cob, it is necessary to reconsider, if there will be tasks, the interval will be subdivided by the area assigned to the functions.
- Significantly all the points, which are located in the required interval, in which there is no first change. Sound the stench of functions, de the argument is placed at the sign of the module, and for state functions with a fractionally rational indicator. As well as the points of vіdsutnі, you can go to the offensive croc.
- Now it is significant, yakі stationary points to spend up to the given interval. The back of the head is equal to 0, it is equal to the same and the root is taken. If we can’t find a suitable stationary point, or the stench doesn’t take intervals from tasks, then we’ll immediately move on to further tasks. Їx determines the interval.
- How can I look at the interval [a; b) , then you need to calculate the value of the function at the point x = a i one-way between lim x → b - 0 f (x) .
- If we look at the interval (a; b], then we need to calculate the value of the function at the point x = b and one-sided boundary lim x → a + 0 f (x).
- If we look at the interval (a; b), then we need to calculate one-sided inter lim x → b - 0 f (x), lim x → a + 0 f (x).
- How can I look at the interval [a; + ∞) , then you need to calculate the value of the point x = a i between the plus inconsistencies lim x → + ∞ f (x) .
- How the interval looks like (- ∞ ; b ) , the value at the point x = b і is calculated at the minus infinity lim x → - ∞ f (x) .
- Yakscho - ∞; b , then one-sided between lim x → b - 0 f (x) and between minus inconsistency lim x → - ∞ f (x)
- Yakscho w - ∞; + ∞ , then we take into account the minus i plus inconsistencies lim x → + ∞ f (x), lim x → - ∞ f (x).
- For example, it is necessary to grow visnovok on the basis of taking away the value of the function and between. There are no options here. So, even though it is a one-sided boundary between the most important minus of inconsistency or plus of inconsistency, then I realized that it is impossible to say anything about the least and most important functions. Below we will analyze one typical butt. Detailed descriptions will help you understand what is up to what. If necessary, you can turn to small pieces 4 - 8 at the first part of the material.
butt 2 Umov: given a function y = 3 e 1 x 2 + x - 6 - 4 . Calculate the largest and smallest values in intervals - ∞ ; - 4, - ∞; - 3, (-3; 1], (-3; 2), [1; 2), 2; +∞, [4; +∞).
Solution
We are aware of the scope of the assigned function. At the banner of the fraction there is a square trinomial, which is not guilty of turning to 0:
x 2 + x - 6 = 0 D = 1 2 - 4 1 (- 6) = 25 x 1 = - 1 - 5 2 = - 3 x 2 = - 1 + 5 2 = 2 ⇒ D (y) : x ∈ (- ∞ ; - 3) ∪ (- 3 ; 2) ∪ (2 ; + ∞)
We took away the area of designated function, until all appointments lie within the interval.
Now we can see the differentiation of functions and take them away:
y "= 3 e 1 x 2 + x - 6 - 4" = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " == 3 e 1 x 2 + x - 6 1” x 2 + x - 6 - 1 x 2 + x - 6” (x 2 + x - 6) 2 = - 3 (2 x + 1) e 1 x 2 + x - 6 x 2 + x - 6 2
Otzhe, pokhіdnі її ї ї іsnuyut іnіy oblastі її її znachennya.
Let's move on to the significance of stationary points. Pokhіdna functions go down to 0 at x = - 1 2 . This is a stationary point, as it is in the intervals (-3; 1] and (-3; 2).
We calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ], as well as the interval for the minus inconsistency:
y (- 4) \u003d 3 e 1 (- 4) 2 + (- 4) - 6 - 4 \u003d 3 e 1 6 - 4 ≈ - 0. 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1
Oskіlki 3 e 1 6 - 4 > - 1 , so m a x y x ∈ (- ∞ ; - 4 ) = y (- 4) = 3 e 1 6 - 4. This does not give us the ability to uniquely determine the least value of the function. the growth of the visnovok, which is below the fringe - 1, the scaling of the function itself to its value is approached asymptotically by the minus of the inconsistency.
The peculiarities of another interval are those that are in the new one, there are no stable points of the same sharp boundary. Also, neither the largest nor the smallest value of the function can be calculated. Having marked the boundary by minus inconsistency with the argument up to - 3 on the left side, we take only the interval value:
lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 = 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
Mean the values of the function will be expanded in the interval - 1; +∞
In order to know the largest value of the function for the third interval, it is significant that the value of the stationary point is x = - 1 2 , so x = 1 . Also, we need to know the one-sided boundary for that vipadka, if the argument is pragne up to - 3 from the right side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e 4 25 - 4 ≈ - 1 . 444 y (1) = 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1 . 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 - 3 + 0 + 3 (-3 + 0 - 2) - 4 = = 3 e 1 (- 0) - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
We have seen that the largest value of the function will be at the stationary point m a x y x ∈ (3 ; 1 ] = y - 1 2 = 3 e - 4 25 - 4. – tse subsoiling from the bottom up to - 4 .
For the interval (- 3 ; 2), we take the results of the forward calculation and once again, we praise, why the one-sided boundary is better when exercising up to 2 from the left side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e - 4 25 - 4 ≈ - 1 . 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 = = 3 e 1 - 0 - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
Then, m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4 and the smallest value cannot be calculated, and the value of the function is subdivided from below by the number - 4 .
Depending on what we had in the two previous calculations, we can confirm that on the interval [1; 2) the largest value of the function is accepted at x = 1, but it is impossible to know the least.
On the interval (2 ; + ∞), the function reaches neither the largest nor the smallest value, that is. won't take the value of the interval - 1; +∞.
lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
Having calculated why the value of the function is more important for x = 4 , it is clear that m a x y x ∈ [ 4 ; + ∞) = y (4) = 3 e 1 14 - 4 i the function on plus infinity is set to asymptotically approach the straight line y = - 1 .
Porіvnyaєmo those that we have seen in the skin count, with the schedule of the assigned function. A small asymptote is shown by a dotted line.
That's all we wanted to know about the significance of the largest and smallest functions. These sequences, which we have brought, will help you to complete the necessary calculations as quickly as possible and simply. But remember, that you often fold the back of the head, at some intervals the function changes, and at some increase, after which you can work farther away. So it is possible to more accurately determine the most and least of the functions and to reduce the results.
How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter
Problem Statement 2:
The function is given, assigned and without interruption to the singing interval. It is necessary to know the largest (least) value of the function for each space.
Theoretical foundations.
Theorem (Other theorem of Weierstrass):
As the function is assigned and without interruption in a closed space, it will reach its largest and smallest value.
The function can reach its highest and lowest values either at the inner points of the gap or at the other boundaries. We illustrate all possible options.
Explanation:
1) The function reaches its greatest value on the left interspace at the point, and its smallest value on the right interspace at the point.
2) The function reaches its largest value at the point (the point to the maximum), and its smallest value at the right interval at the point.
3) The function reaches its largest value on the left interspace at the point, and its smallest value at the point (the whole point is the minimum).
4) The function has become on promizhku, tobto. won't reach its minimum and maximum value at any point in the interim, moreover, the minimum and maximum value are equal to each other.
5) The function reaches its largest value at a point, and its smallest value at a point (before those, that the function can have a maximum, and a minimum).
6) The function reaches its largest value at the point (the point is the maximum), and its smallest value is at the point (the point is the minimum).
Respect:
"Maximum" and "maximum value" - different speeches. The reason is clear from the assignment to the maximum of that intuitively logical understanding of the phrase “maximum value”.
Algorithm for decoupling tasks 2.
4) Select from the subtract the value of the largest (least) and write down the difference.
Example 4:
Calculate the largest and smallest value of the function  on the vіdrіzku.
Solution:
1) Know the appropriate functions.
2) Find stationary points (i points, suspected for extremum), virishivshi equalization. Turn your respect to the points, in which there is no two-sided end-of-life.
3) Calculate the value of the function at stationary points and at the boundaries of the interval.
4) Select from the subtract the value of the largest (least) and write down the difference.
The function on which window reaches its greatest value at the point with coordinates.
The function on which view reaches the smallest value at the point with the coordinates.
The correctness of the calculation can be reconsidered, marveling at the graph of the completed function.
Respect: The largest value of the function is at the point of maximum, and the smallest is between the points.
Okremy is a jerk.
It is acceptable, it is necessary to know the maximum and minimum value of the current function for the wind. Following the violation of the first paragraph of the algorithm, tobto. It becomes clear that, for example, it does not take on any more negative meanings in every respect. Remember, if it is negative, then the function changes. We took away that the function changes in every respect. This situation is shown on graph No. 1 on the cob of the stat.
The function will change to a different one, tobto. she has no extremum point. It can be seen from the image that the least value of the function is taken on the right side of the window, and the most value is on the left. if it is similar to the wind, it is everywhere positive, then the function grows. The least value is on the left side of the window, the most - on the right.
At this article, I will tell you about algorithm for finding the largest and smallest values functions, point of minimum and maximum. We need theories tableі rules of differentiation. All the same at this table: Algorithm to search for the largest and smallest value. I can explain more clearly on a specific example. Let's take a look: Butt: Find the largest value of the function y=x^5+20x^3–65x on the reverse [–4;0]. Krok 1. Let's go away. Y" = (x^5+20x^3–65x)" = 5x^4 + 20*3x^2 - 65 = 5x^4 + 60x^2 - 65 Krok 2 We know the extremum points. Krapkoy extremum we call such points, for which the function reaches its highest or lowest value. To know the extremum points, you need to equate a similar function to zero (y" = 0) 5x^4 + 60x^2 - 65 = 0 Now it is clear that the square is equal and that the root is known and our points of extremum. I will untie this equalization with a replacement t = x^2, then 5t^2 + 60t - 65 = 0. Quickly equalize by 5, take: t^2 + 12t - 13 = 0 D = 12^2 - 4 * 1 * (-13) = 196 T_(1) = (-12 + sqrt(196))/2 = (-12 + 14)/2 = 1 T_(2) = (-12 - sqrt(196))/2 = (-12 - 14)/2 = -13 Robimo zavorotnu zamіnu x^2 = t: X_(1 and 2) = ±sqrt(1) = ±1
x_(3 і 4) = ±sqrt(-13) Together: x_(1) = 1 і x_(2) = -1 - і є our extremum points. Krok 3 The most important is the least significant. Substitution method. For the mind, we were given a vіdrіzok [b] [–4; 0]. The point x=1 must not enter this branch. Otzhe її we do not see. In addition to the points x=-1, we also need to look at the left and right between our vіdrіzka, then the points -4 and 0. For which we represent all the three points at the output function. Respect the vihіdnu - tse that, as given in the mind (y=x^5+20x^3–65x), deyakі start imposing at the pokhіdnu... Y(-1) = (-1)^5 + 20*(-1)^3 - 65*(-1) = -1 - 20 + 65 = [b]44
y(0) = (0)^5 + 20*(0)^3 - 65*(0) = 0
y(-4) = (-4)^5 + 20*(-4)^3 - 65*(-4) = -1024 - 1280 + 260 = -2044 Mean the greatest value of the function [b]44 and it reaches out to the point [b]-1, as it is called the maximum point of the function on the top [-4; 0]. We vyrishili and otrimali vіdpovіd, mi good fellows, you can relax. Ale stop! You don't know how good y(-4) is supposed to be? In the minds of an obedient hour, it’s better to speed up in a different way, I call yoga like this: Through the passages of signs. To know the number of gaps for a casual function, tobto our b_kvadratnogo equal. I work like this. Small straightening of the vіdrіzok. I set the dots: -4, -1, 0, 1. Do not be surprised at those that 1 is not included in the tasks of the entries, її all one should be assigned in order to correctly designate the gaps of familiarity. Let's take it, maybe the number is more than 1, 100 is acceptable, we can think of it in our biquadratic equal 5 (100) ^ 4 + 60 (100) ^ 2 - 65. Navigate nothing crim becomes obvious that at point 100 the function of ma є plus sign. And that means that for promishes from 1 to 100 won there is a plus sign. When passing through 1 (we are right-handed left), the function will change the sign to minus. When passing through the point 0, the function saves its sign, the shards are less than the boundary of the vіdrіzka, and the chi is not the root of the equal. When passing through -1, the function will change the plus sign again. 
From theory, we know what the functions are there (and we didn’t arm ourselves for it) change sign from plus to minus (point -1 in our vipad) function available its local maximum (y(-1)=44, as it was buffed earlier) on this vіdrіzku (it is logically more reasonable, the function has ceased to grow, the shards have reached their maximum and have begun to change). Obviously, there are some useful functions change sign from minus to plus, reachable local minimum of the function. So, so, we also know the point of the local minimum 1, and y(1) - the minimum value of the function on the top, it is permissible from -1 to +∞. Give great respect, which is only a local minimum, then a minimum on a singing wind. So how is the actual (global) minimum of the function reachable here, -∞. At a glance, the first way is simpler theoretically, and the other is simpler from a glance of arithmetic actions, but richer from a glance of theory. Even if the function does not change the sign when passing through the equal root, you can sometimes get lost with both local and global maxima and minima, if you want to get good at it, then you plan to enter ati to technical VNZ (and for why else do you have a profile EDI that virishuvati tse zavdannya). Ale practice and less practice once and for all to teach you how to do this task. And you can train on our website. Axis. Yakshcho vynikli yakіs pitanya, but schos unreasonably - obov'yazkovo energize. I am glad to see you and will make changes, adding to the article. Remember, mi robimo this site at once!
From a practical point of view, the most interesting is the variation of the similar value of the largest and smallest value of the function. Why is it related? Maximization of profit, minimization of vitrate, determination of the optimal cost of installation... It seems that in rich spheres of life it is necessary to violate the task of optimizing whatever parameters. And the goal is to specify the value of the largest and smallest value of the function.
Next, designate which is the largest and least important function, sound on the interval X, which is the entire area of function assignment or a partial area of assignment. The X interval itself can be a different, critical interval  , inexhaustible intercourse.
In this article, we speak about the significance of the largest and smallest values of the explicitly given function of one variable y=f(x) .
Navigation on the side.
The most important and least important functions are designations, illustrations.
The stylus is sharpened on the main designations.
The highest value of the function  , what for be-whom  fair nerіvnіst.
The smallest values of the function y=f(x) for interval X  , what for be-whom fair nerіvnіst.
The value of the value is intuitively reasonable: the largest (least) value of the function is the largest (smaller) value on the analyzed interval with the abscissa.
Stationary points- the value of the argument, for some of them, the functions turn to zero.
Why do we need stationary points with the highest and lowest values? Fermat's theorem gives evidence. From the point of the theorem, it is obvious that as a function, as a differentiation, there is an extremum (local minimum or local maximum) at a real point, then this point is stationary. In this way, the function often takes its largest (least) value on the interval X at one of the stationary points of this interval.
Also, often the smallest value of a function can be taken at points that do not have the first similar function, but the function itself is assigned.
Let's look at one of the widest data on this topic: "What can you ever calculate the most (least) value of a function"? Don't wait. Sometimes between intervals of X are zbіgayutsya from the boundaries of the area assigned to the function, or the interval of X is not limited. And the deacons of the function on the incompatibility and on the borders of the region of appointment can be as infinitely large, so infinitely small value. In these cases, nothing can be said about the most and least important functions.
For clarity, I'll give a graphic illustration. Look at the little ones - and richly clear up.
On the vіdrіzka
 On the first baby, the function takes the most (max y) and least (min y) values at stationary points, which are in the middle of the circle [-6; 6].
Let's look at the vipadok, images of another little one. Let's change it to . For this example, the least value of the function is reached at the stationary point, and the most - at the point with the abscissa, which shows the right inter-interval.
On little No. 3, the boundary points of the cross [-3; 2] are the abscissa points that correspond to the largest and smallest value of the function.
At a specified interval
 On the fourth small, the function takes the most (max y ) and the least (min y ) values of stationary points, which are in the middle of the open interval (-6; 6).
At the interval about the most significant, no changes are possible.
On inconsistency
 In the butt, presented on the soma baby, the function takes on the largest value (max y) at the stationary point with the abscissa x=1, and the smallest value (min y) is reached on the right interinterval. At the minus of inconsistency, the values of the function asymptotically approach up to y=3.
On the interval, the function does not reach the smallest, nor the largest value. When the value of the function is right-handed up to x=2, the value of the function is assumed to be minus inconsistency (straight line x=2 is the vertical asymptote), and when the abscissa is correct up to plus inconsistency, the value of the function asymptotically approaches up to y=3. A graphic illustration of the butt of the small butt No. 8.
Algorithm for finding the largest and smallest values of a non-permanent function on a winder.Let's write down an algorithm that allows you to know the largest and smallest values of the function on the input.
- We know the scope of the assigned function and it is reverified, so that the whole vdrіzok can be removed from it.
- We know all the points, in which the first is not the case, and in which the first is lost, and in which they are near the wind (sound in such a way the points are picked up for functions with an argument under the modulus sign and for stationary functions with a fractional-rational exponent). Since there are no such points, we pass to the offensive point.
- All stationary points are visible, which are in the winds. For whom, equating it to zero, it is better to omit equal and choose the same root. There are not many stationary points, but you can’t waste them at the windbreaks, let’s move on to the offensive point.
- Calculating the value of the function at selected stationary points (such as є), at points, which do not have the first line (such as є), and also at x=a and x=b.
- To take away the value of the function, choose the most and least - they will be the highest and least value of the function, obviously.
Let's analyze the algorithm when we apply the value of the highest and the lowest value of the function to the top.
butt.
Find the highest and lowest value of a function - on the vіdrіzku;
- to the reverse [-4;-1].
Solution.
The scope of the function is the impersonal real numbers, the cream of zero, tobto. Offenses are taken from the designated area.
We know similar functions by:
Obviously, similar functions exist at all points in the intersection and [-4;-1].
Stationary points are significantly more equal. The only real root is є x=2. Tsya stationary point is consumed at the first vіdrіzok.
For the first type, the value of the function at the ends of the cut and at the stationary point is calculated, so at x=1, x=2 and x=4:
Same, most important function  reachable at x=1 and the smallest value  - When x = 2.
For another way, the value of the function is calculated only on the ends of the contraction [-4;-1] (scaling wines do not avenge the same stationary point):
Solution.
Let's start with the area of assigned function. The square trinomial at the banner of the fraction is not guilty of turning to zero:
It is easy to overestimate that all intervals should be considered to lie in the area of assigned function.
Prodifferentiation function:
Obviously, the function is similar in all areas.
We know the stationary points. Pokhіdna turns to zero at . This stationary point is consumed in the interval (-3; 1) and (-3; 2).
And now you can take the results from the function graph at the skin point. The blue dotted lines show the asymptotes.
 On which you can finish from the values of the largest and the smallest value of the function. Algorithms, developed by these stats, allow you to take the results with a minimum of diy. However, on the back of the hand, the increase in the increase and the change in the function and only a little more work of the visnovka about the greatest and least importance of the function on the same interval. This gives a clearer picture of the sum total of results.
Nove
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