What is an extremum of a function and what is the necessary condition for an extremum?
The extremum of a function is the maximum and minimum of the function.
Necessary condition The maximum and minimum (extremum) of a function are as follows: if the function f(x) has an extremum at the point x = a, then at this point the derivative is either zero, or infinite, or does not exist.
This condition is necessary, but not sufficient. The derivative at the point x = a can go to zero, infinity, or not exist without the function having an extremum at this point.
What's it like sufficient condition extremum of the function (maximum or minimum)?
First condition:
If, in sufficient proximity to the point x = a, the derivative f?(x) is positive to the left of a and negative to the right of a, then at the point x = a the function f(x) has maximum
If, in sufficient proximity to the point x = a, the derivative f?(x) is negative to the left of a and positive to the right of a, then at the point x = a the function f(x) has minimum provided that the function f(x) here is continuous.
Instead, you can use the second sufficient condition for the extremum of a function:
Let at the point x = a the first derivative f?(x) vanish; if the second derivative f??(a) is negative, then the function f(x) has a maximum at the point x = a, if it is positive, then it has a minimum.
What is the critical point of a function and how to find it?
This is the value of the function argument at which the function has an extremum (i.e. maximum or minimum). To find it you need find the derivative function f?(x) and, equating it to zero, solve the equation f?(x) = 0. The roots of this equation, as well as those points at which the derivative of this function does not exist, are critical points, i.e., values of the argument at which there can be an extremum. They can be easily identified by looking at derivative graph: we are interested in those values of the argument at which the graph of the function intersects the abscissa axis (Ox axis) and those at which the graph suffers discontinuities.
For example, let's find extremum of a parabola.
Function y(x) = 3x2 + 2x - 50.
Derivative of the function: y?(x) = 6x + 2
Solve the equation: y?(x) = 0
6x + 2 = 0, 6x = -2, x = -2/6 = -1/3
IN in this case the critical point is x0=-1/3. It is with this argument value that the function has extremum. To him find, substitute the found number in the expression for the function instead of “x”:
y0 = 3*(-1/3)2 + 2*(-1/3) - 50 = 3*1/9 - 2/3 - 50 = 1/3 - 2/3 - 50 = -1/3 - 50 = -50.333.
How to determine the maximum and minimum of a function, i.e. its largest and smallest values?
If the sign of the derivative when passing through the critical point x0 changes from “plus” to “minus”, then x0 is maximum point; if the sign of the derivative changes from minus to plus, then x0 is minimum point; if the sign does not change, then at point x0 there is neither a maximum nor a minimum.
For the example considered:
We take an arbitrary value of the argument to the left of the critical point: x = -1
At x = -1, the value of the derivative will be y?(-1) = 6*(-1) + 2 = -6 + 2 = -4 (i.e. the sign is “minus”).
Now we take an arbitrary value of the argument to the right of the critical point: x = 1
At x = 1, the value of the derivative will be y(1) = 6*1 + 2 = 6 + 2 = 8 (i.e. the sign is “plus”).
As you can see, the derivative changed sign from minus to plus when passing through the critical point. This means that at the critical value x0 we have a minimum point.
The greatest and nai lower value functions on the interval(on a segment) are found using the same procedure, only taking into account the fact that, perhaps, not all critical points will lie within the specified interval. Those critical points that are outside the interval must be excluded from consideration. If there is only one critical point inside the interval, it will have either a maximum or a minimum. In this case, to determine the largest and smallest values of the function, we also take into account the values of the function at the ends of the interval.
For example, let's find the largest and smallest values of the function
y(x) = 3sin(x) - 0.5x
at intervals:
So, the derivative of the function is
y?(x) = 3cos(x) - 0.5
We solve the equation 3cos(x) - 0.5 = 0
cos(x) = 0.5/3 = 0.16667
x = ±arccos(0.16667) + 2πk.
We find critical points on the interval [-9; 9]:
x = arccos(0.16667) - 2π*2 = -11.163 (not included in the interval)
x = -arccos(0.16667) – 2π*1 = -7.687
x = arccos(0.16667) - 2π*1 = -4.88
x = -arccos(0.16667) + 2π*0 = -1.403
x = arccos(0.16667) + 2π*0 = 1.403
x = -arccos(0.16667) + 2π*1 = 4.88
x = arccos(0.16667) + 2π*1 = 7.687
x = -arccos(0.16667) + 2π*2 = 11.163 (not included in the interval)
We find the values of the function at critical values argument:
y(-7.687) = 3cos(-7.687) - 0.5 = 0.885
y(-4.88) = 3cos(-4.88) - 0.5 = 5.398
y(-1.403) = 3cos(-1.403) - 0.5 = -2.256
y(1.403) = 3cos(1.403) - 0.5 = 2.256
y(4.88) = 3cos(4.88) - 0.5 = -5.398
y(7.687) = 3cos(7.687) - 0.5 = -0.885
It can be seen that on the interval [-9; 9] highest value the function has at x = -4.88:
x = -4.88, y = 5.398,
and the smallest - at x = 4.88:
x = 4.88, y = -5.398.
On the interval [-6; -3] we have only one critical point: x = -4.88. The value of the function at x = -4.88 is equal to y = 5.398.
Find the value of the function at the ends of the interval:
y(-6) = 3cos(-6) - 0.5 = 3.838
y(-3) = 3cos(-3) - 0.5 = 1.077
On the interval [-6; -3] we have the greatest value of the function
y = 5.398 at x = -4.88
smallest value -
y = 1.077 at x = -3
How to find the inflection points of a function graph and determine the convex and concave sides?
To find all the inflection points of the line y = f(x), you need to find the second derivative, equate it to zero (solve the equation) and test all those values of x for which the second derivative is zero, infinite or does not exist. If, when passing through one of these values, the second derivative changes sign, then the graph of the function has an inflection at this point. If it doesn’t change, then there is no bend.
The roots of the equation f? (x) = 0, as well as possible discontinuity points of the function and the second derivative, divide the domain of definition of the function into a number of intervals. The convexity on each of their intervals is determined by the sign of the second derivative. If the second derivative at a point on the interval under study is positive, then the line y = f(x) is concave upward, and if negative, then downward.
How to find the extrema of a function of two variables?
To find the extrema of the function f(x,y), differentiable in the domain of its specification, you need:
1) find the critical points, and for this - solve the system of equations
fх? (x,y) = 0, fу? (x,y) = 0
2) for each critical point P0(a;b) investigate whether the sign of the difference remains unchanged
for all points (x;y) sufficiently close to P0. If the difference remains positive sign, then at point P0 we have a minimum, if negative, then we have a maximum. If the difference does not retain its sign, then there is no extremum at point P0.
The extrema of the function are determined similarly for more arguments.
Which carbonated soft drinks clean surfaces?
There is an opinion that the carbonated soft drink Coca-Cola can dissolve meat. But, unfortunately, there is no direct evidence of this. On the contrary, there are affirmative facts confirming that meat left in the Coca-Cola drink for two days changes in consumer properties and does not disappear anywhere.
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How to treat neurosis
Neurosis (Novolat. neurosis, comes from the ancient Greek νε?ρον - nerve; synonyms - psychoneurosis, neurotic disorder) - in the clinic: a collective name for a group of functional psychogenic reversible disorders that tend to persist
What is aphelion
Apocenter is the point in the orbit at which a body revolving in an elliptical orbit around another body reaches its maximum distance from the latter. At the same point, according to Kepler's second law, the speed orbital movement becomes minimal. The apocenter is located at a point diametrically opposite to the periapsis. In special cases, it is customary to use special terms:
What is mamon
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How to determine the convexity and concavity intervals of a function graph
What is an extremum of a function and what is the necessary condition for an extremum? The extremum of a function is the maximum and minimum of the function. The necessary condition for the maximum and minimum (extremum) of a function is the following: if the function f(x) has an extremum at the point x = a, then at this point the derivative is either zero, infinite, or does not exist. This condition is necessary, but not sufficient. Derivative in t
The largest (smallest) value of a function is the largest (smallest) accepted value of the ordinate on the considered interval.
To find the largest or smallest value of a function you need to:
- Check which stationary points are included in a given segment.
- Calculate the value of the function at the ends of the segment and at stationary points from step 3
- Select the largest or smallest value from the results obtained.
To find the maximum or minimum points you need to:
- Find the derivative of the function $f"(x)$
- Find stationary points by solving the equation $f"(x)=0$
- Factor the derivative of a function.
- Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the resulting intervals, using the notation in step 3.
- Find the maximum or minimum points according to the rule: if at a point the derivative changes sign from plus to minus, then this will be the maximum point (if from minus to plus, then this will be the minimum point). In practice, it is convenient to use the image of arrows on intervals: on the interval where the derivative is positive, the arrow is drawn upward and vice versa.
Table of derivatives of some elementary functions:
| Function | Derivative |
| $c$ | $0$ |
| $x$ | $1$ |
| $x^n, n∈N$ | $nx^(n-1), n∈N$ |
| $(1)/(x)$ | $-(1)/(x^2)$ |
| $(1)/x(^n), n∈N$ | $-(n)/(x^(n+1)), n∈N$ |
| $√^n(x), n∈N$ | $(1)/(n√^n(x^(n-1)), n∈N$ |
| $sinx$ | $cosx$ |
| $cosx$ | $-sinx$ |
| $tgx$ | $(1)/(cos^2x)$ |
| $ctgx$ | $-(1)/(sin^2x)$ |
| $cos^2x$ | $-sin2x$ |
| $sin^2x$ | $sin2x$ |
| $e^x$ | $e^x$ |
| $a^x$ | $a^xlna$ |
| $lnx$ | $(1)/(x)$ |
| $log_(a)x$ | $(1)/(xlna)$ |
Basic rules of differentiation
1. The derivative of the sum and difference is equal to the derivative of each term
$(f(x) ± g(x))′= f′(x)± g′(x)$
Find the derivative of the function $f(x) = 3x^5 – cosx + (1)/(x)$
The derivative of the sum and difference is equal to the derivative of each term
$f′(x)=(3x^5)′–(cosx)′+((1)/(x))"=15x^4+sinx-(1)/(x^2)$
2. Derivative of the product.
$(f(x)∙g(x))′=f′(x)∙g(x)+f(x)∙g(x)′$
Find the derivative $f(x)=4x∙cosx$
$f′(x)=(4x)′∙cosx+4x∙(cosx)′=4∙cosx-4x∙sinx$
3. Derivative of the quotient
$((f(x))/(g(x)))"=(f^"(x)∙g(x)-f(x)∙g(x)")/(g^2(x) )$
Find the derivative $f(x)=(5x^5)/(e^x)$
$f"(x)=((5x^5)"∙e^x-5x^5∙(e^x)")/((e^x)^2)=(25x^4∙e^x- 5x^5∙e^x)/((e^x)^2)$
4. Derivative complex function is equal to the product of the derivative of the external function and the derivative of the internal function
$f(g(x))′=f′(g(x))∙g′(x)$
$f′(x)=cos′(5x)∙(5x)′= - sin(5x)∙5= -5sin(5x)$
Find the minimum point of the function $y=2x-ln(x+11)+4$
1. Let's find ODZ functions: $x+11>0; x>-11$
2. Find the derivative of the function $y"=2-(1)/(x+11)=(2x+22-1)/(x+11)=(2x+21)/(x+11)$
3. Find stationary points by equating the derivative to zero
$(2x+21)/(x+11)=0$
A fraction is equal to zero if the numerator equal to zero, and the denominator is not zero
$2x+21=0; x≠-11$
4. Let's draw a coordinate line, place stationary points on it and determine the signs of the derivative in the resulting intervals. To do this, substitute any number from the rightmost region into the derivative, for example, zero.
$y"(0)=(2∙0+21)/(0+11)=(21)/(11)>0$
5. At the minimum point, the derivative changes sign from minus to plus, therefore, the point $-10.5$ is the minimum point.
Answer: $-10.5$
Find the greatest value of the function $y=6x^5-90x^3-5$ on the segment $[-5;1]$
1. Find the derivative of the function $y′=30x^4-270x^2$
2. Equate the derivative to zero and find stationary points
$30x^4-270x^2=0$
We'll take it out common multiplier$30x^2$ in brackets
$30x^2(x^2-9)=0$
$30x^2(x-3)(x+3)=0$
Let's equate each factor to zero
$x^2=0 ; x-3=0; x+3=0$
$x=0;x=3;x=-3$
3. Select stationary points that belong to this segment $[-5;1]$
The stationary points $x=0$ and $x=-3$ suit us
4. Calculate the value of the function at the ends of the segment and at stationary points from step 3
And to solve it you will need minimal knowledge of the topic. The next one ends academic year, everyone wants to go on vacation, and to bring this moment closer, I’ll get straight to the point:
Let's start with the area. The area referred to in the condition is limited closed set of points on a plane. For example, the set of points bounded by a triangle, including the WHOLE triangle (if from borders“prick out” at least one point, then the region will no longer be closed). In practice, there are also areas that are rectangular, circular, and slightly larger. complex shapes. It should be noted that in theory mathematical analysis strict definitions are given limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and now nothing more is needed.
A flat area is standardly denoted by the letter , and, as a rule, is specified analytically - by several equations (not necessarily linear); less often inequalities. Typical turn of phrase: "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? You need to draw all the listed lines (in this case 3 straight) and analyze what happened. The searched area is usually lightly shaded, and its border is marked with a thick line: 
The same area can also be set linear inequalities: , which for some reason are often written as an enumerated list rather than system.
Since the boundary belongs to the region, then all inequalities, of course, lax.
And now the essence of the task. Imagine that the axis comes out straight towards you from the origin. Consider a function that continuous in each area point. The graph of this function represents some surface, And a little happiness is 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 does not matter. And the following is important: according to Weierstrass's theorems, continuous V limited closed area the function reaches its greatest value (the “highest”) and the least (the “lowest”) values that need to be found. Such values are achieved or V stationary points, belonging to the regionD , or at points that lie on the border of this area. This leads to a simple and transparent solution algorithm:
Example 1
In limited closed area
Solution: First of all, you need to depict the area in the drawing. Unfortunately, it is technically difficult for me to do interactive model task, and therefore I will immediately present the final illustration, which depicts all the “suspicious” points found during the study. They are usually listed one after the other as they are discovered: 
Based on the preamble, it is convenient to break the decision into two points:
I) Find stationary points. This is a standard action that we performed repeatedly in class. about extrema of several variables:
Found stationary point belongs areas: (mark it on the drawing), which means we should calculate the value of the function at a given point:
- as in the article The largest and smallest values of a function on a segment, important results I'll put it in bold. It is convenient to trace them in a notebook with a pencil.
Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at a point the function reaches, for example, local minimum , then this DOES NOT MEAN that the resulting value will be minimal throughout the region (see the beginning of the lesson about unconditional extremes) .
What to do if the stationary point does NOT belong to the region? Almost nothing! It should be noted that and move on to the next point.
II) We explore the border of the region.
Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subsections. But it’s better not to do it anyhow. From my point of view, it is more advantageous to first consider the segments parallel coordinate axes, and first of all, those lying on the axes themselves. To grasp the entire sequence and logic of actions, try to study the ending “in one breath”:
1) Let's deal with the bottom side of the triangle. To do this, substitute directly into the function:
Alternatively, you can do it like this: 
Geometrically this means that coordinate plane (which is also given by the equation)"carves" out of surfaces a "spatial" parabola, the top of which immediately comes under suspicion. Let's find out where is she located:
– the resulting value “fell” into the area, and it may well turn out that at the point (marked on the drawing) the function reaches the largest or smallest value in the entire region. One way or another, let's do the calculations:
The other “candidates” are, of course, the ends of the segment. Let's calculate the values of the function at points 
(marked on the drawing):
Here, by the way, you can perform an oral mini-check using a “stripped-down” version: 
2) For research right side we substitute the triangle into the function and “put things in order”:
Here we will immediately perform a rough check, “ringing” the already processed end of the segment:
, Great.
The geometric situation is related to the previous point:
– the resulting value also “came into the sphere of our interests,” which means we need to calculate what the function at the appeared point is equal to:
Let's examine the second end of the segment:
Using the function 
, let's perform a control check: 
3) Probably everyone can guess how to explore the remaining side. We substitute it into the function and carry out simplifications:
Ends of the segment 
have already been researched, but in the draft we 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 anything interesting inside the segment:
- There is! Substituting the straight line into the equation, we get the ordinate of this “interestingness”:
We mark a point on the drawing and find the corresponding value of the function:
Let’s check the calculations using the “budget” version 
:
, order.
And the final step: We CAREFULLY look through all the “bold” numbers, I recommend that beginners even make a single list: 
from which we select the largest and smallest values. Answer Let's write down in the style of the problem of finding the largest and smallest values of a function on a segment:
Just in case, I'll comment again geometric meaning result:
- here is the most high point surfaces in the area ;
- here is the most low point surfaces in the area.
In the analyzed task, we identified 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum "research set" consists of three points. This happens when the function, for example, specifies plane– it is completely clear that there are no stationary points, and the function can reach its maximum/smallest values only at the vertices of the triangle. But there are only one or two similar examples - usually you have to deal with some kind of surface of 2nd order.
If you try to solve such tasks a little, then the triangles can make your head spin, and that’s why I prepared for you unusual examples so that it becomes square :))
Example 2
Find the largest and smallest values of a function 
in a closed area bounded by lines
Example 3
Find the largest and smallest values of a function in a limited closed region.
Special attention Pay attention to the rational order and technique of studying the boundary of the region, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it any way you like, but in some problems, for example, in Example 2, there is every chance of making your life much more difficult. Approximate sample 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 of the 1st example:
– In the first step, we build an area, it is advisable to shade it and highlight the border with a bold line. During the solution, points will appear that need to be marked on the drawing.
– Find stationary points and calculate the values of the function only in those of them that belong to the region. We highlight the resulting values in the text (for example, circle them with a pencil). If a stationary point does NOT belong to the region, then we mark this fact with an icon or verbally. If stationary points not at all, then we draw a written conclusion that they are absent. In any case, this point cannot be skipped!
– We are exploring the border of the region. First, it is beneficial to understand the straight lines that are parallel to the coordinate axes (if there are any at all). We also highlight the function values 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 it!
– From the selected numbers, select the largest and smallest values and give the answer. Sometimes it happens that a function reaches such values at several points at once - in this case, all these points should be reflected in the answer. Let, for example, 
and it turned out that this is the smallest value. Then we write down that
The final examples are dedicated to others useful ideas which will be useful in practice:
Example 4
Find the largest and smallest values of a function in a closed region 
.
I have retained the author's formulation, in which the region is given in the form of a double inequality. This condition can be written equivalent system or in a more traditional form for this task: 
I remind you that with nonlinear we encountered inequalities on, and if you do not understand the geometric meaning of the notation, then please do not delay and clarify the situation right now;-)
Solution, as always, begins with constructing an area that represents a kind of “sole”: 
Hmm, sometimes you have to chew not only the granite of science...
I) Find stationary points: 
The system is an idiot's dream :) 
A stationary point belongs to the region, namely, lies on its boundary.
And so, it’s okay... the lesson went well - this is what it means to drink the right tea =)
II) We explore the border of the region. Without further ado, let's start with the x-axis:
1) If , then
Let's find where the vertex of the parabola is:
– appreciate such moments – you “hit” right to the point from which everything is already clear. But we still don’t forget about checking: 
Let's calculate the values of the function at the ends of the segment: 
2) Let’s deal with the lower part of the “sole” “in one sitting” - without any complexes we substitute it into the function, and we will only be interested in the segment:
Control:
This already brings some excitement to the monotonous driving along the knurled track. Let's find critical points:
Let's decide quadratic equation, do you remember anything else about this? ...However, remember, of course, otherwise you wouldn’t be reading these lines =) If in the two previous examples calculations in decimals(which, by the way, is rare), then the usual ones await us here common 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 of the function at the found points: 
Check the function yourself.
Now we carefully study the won trophies and write down answer:
These are “candidates”, these are “candidates”!
For independent decision:
Example 5
Find the smallest and largest values of a function 
in a closed area 
Recording from curly braces reads like this: “a set of points such that.”
Sometimes in similar examples use Lagrange multiplier method, but there is unlikely to be a real need to use it. So, for example, if a function with the same area “de” is given, then after substitution into it – with the derivative from 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 more complex cases, where without the Lagrange function (where, for example, is the same equation of a circle) It’s hard to get by – just as it’s hard to get by without a good rest!
Have a good time everyone and see you soon next season!
Solutions and answers:
Example 2: Solution: Let's depict the area in the drawing:
Petite and pretty simple task from the category of those that serve as a life preserver for a floating student. It's mid-July in nature, so it's time to settle down with your laptop on the beach. Early morning started playing sunny bunny theory in order to soon focus on practice, which, despite its claimed ease, contains glass shards in the sand. In this regard, I recommend that you conscientiously consider the few examples of this page. For solutions practical tasks must be able to find derivatives and understand the material of the article Monotonicity intervals and extrema of the function.
First, briefly about the main thing. In the lesson about continuity of function I gave the definition of continuity at a point and continuity at an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on an interval if:
1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.
In the second paragraph we talked about the so-called one-sided continuity functions at a point. There are several approaches to defining it, but I will stick to the line I started earlier:
The function is continuous at the point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: 
. It is continuous at the point left, if defined at a given point and its left-sided limit equal to the value at this point: 
Imagine that green dots- these are the nails on which the magic elastic band is attached:
Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited– a fence at the top, a fence at the bottom, and our product grazes in the paddock. Thus, a function continuous on an interval is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and strictly proven. Weierstrass's first theorem....Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled a graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Really, how do you know what awaits us over the horizon? After all, the Earth was once considered flat, so today even ordinary teleportation requires proof =)
According to Weierstrass's second theorem, continuous on a segmentthe function reaches its accurate top edge and yours exact bottom edge .
The number is also called the maximum value of the function on the segment and are denoted by , and the number is the minimum value of the function on the segment marked .
In our case: 
Note
: in theory, recordings are common 
.
Roughly speaking, the largest value is where the highest point of the graph is, and the smallest value is where the lowest point is.
Important! As already emphasized in the article about extrema of the function, greatest function value And smallest function value – NOT THE SAME, What maximum function And minimum function. So, in the example under consideration, the number is the minimum of the function, but not the minimum value.
By the way, what happens outside the segment? Yes, even a flood, in the context of the problem under consideration, this does not interest us at all. The task only involves finding two numbers 
and that's it!
Moreover, the solution is purely analytical, therefore no need to make a drawing!
The algorithm lies on the surface and suggests itself from the above figure:
1) Find the values of the function in critical points, which belong to this segment.
Catch another bonus: here there is no need to check the sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum doesn't guarantee yet, what is the minimum or maximum value. The demonstration function reaches a maximum and, by the will of fate, the same number is the largest value of the function on the segment. But, of course, such a coincidence does not always take place.
So, in the first step, it is faster and easier to calculate the values of the function at critical points belonging to the segment, without bothering whether there are extrema in them or not.
2) We calculate the values of the function at the ends of the segment.
3) Among the function values found in the 1st and 2nd paragraphs, select the smallest and most big number, write down the answer.
We sit down on the shore blue sea and hit the shallow water with our heels:
Example 1
Find the largest and smallest values of a function on a segment
Solution:
1) Let's calculate the values of the function at critical points belonging to this segment:
Let's calculate the value of the function in the second critical point:
2) Let’s calculate the values of the function at the ends of the segment: 
3) “Bold” results were obtained with exponents and logarithms, which significantly complicates their comparison. For this reason, let’s arm ourselves with a calculator or Excel and calculate approximate values, not forgetting that: 
Now everything is clear.
Answer:
Fractional-rational instance for independent solution:
Example 6
Find the maximum and minimum value functions on an interval
In this article I will talk about how to apply the skill of finding to the study of a function: to find its largest or smallest value. And then we will solve several problems from Task B15 from Open Bank tasks for .
As usual, let's first remember the theory.
At the beginning of any study of a function, we find it
To find the largest or smallest value of a function, you need to examine on which intervals the function increases and on which it decreases.
To do this, we need to find the derivative of the function and examine its intervals of constant sign, that is, the intervals over which the derivative retains its sign.
Intervals over which the derivative of a function is positive are intervals of increasing function.
Intervals over which the derivative of a function is negative are intervals of decreasing function.
1 . Let's solve task B15 (No. 245184)
To solve it, we will follow the following algorithm:
a) Find the domain of definition of the function
b) Let's find the derivative of the function.
c) Let's equate it to zero.
d) Let us find the intervals of constant sign of the function.
e) Find the point at which the function takes on the greatest value.
f) Find the value of the function at this point.
I explain the detailed solution to this task in the VIDEO TUTORIAL:
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2. Let's solve task B15 (No. 282862)
Find the largest value of the function 
on the segment
It is obvious that the function takes the greatest value on the segment at the maximum point, at x=2. Let's find the value of the function at this point:
Answer: 5
3. Let's solve task B15 (No. 245180):
Find the largest value of the function 
1. title="ln5>0">, , т.к. title="5>1">, поэтому это число не влияет на знак неравенства.!}
2. Because according to the domain of definition of the original function title="4-2x-x^2>0">, следовательно знаменатель дроби всегда больще нуля и дробь меняет знак только в нуле числителя.!}
3. The numerator is equal to zero at . Let's check whether the ODZ belongs to the function. To do this, let’s check whether the condition title="4-2x-x^2>0"> при .!}
Title="4-2(-1)-((-1))^2>0">,
this means that the point belongs to the ODZ function
Let's examine the sign of the derivative to the right and left of the point:
We see that the function takes on its greatest value at point . Now let's find the value of the function at:
Remark 1. Note that in this problem we did not find the domain of definition of the function: we only fixed the restrictions and checked whether the point at which the derivative is equal to zero belongs to the domain of definition of the function. This turned out to be sufficient for this task. However, this is not always the case. It depends on the task.
Remark 2. When studying the behavior of a complex function, you can use the following rule:
- if the external function of a complex function is increasing, then the function takes its greatest value at the same point at which internal function takes the greatest value. This follows from the definition of an increasing function: a function increases on interval I if higher value the argument from this interval corresponds to a larger value of the function.
- if the outer function of a complex function is decreasing, then the function takes its greatest value at the same point at which the inner function takes its smallest value . This follows from the definition of a decreasing function: a function decreases on interval I if a larger value of the argument from this interval corresponds to a smaller value of the function
In our example, the external function increases throughout the entire domain of definition. Under the sign of the logarithm there is an expression - quadratic trinomial, which, with a negative leading coefficient, takes the greatest value at the point 
. Next, we substitute this x value into the function equation and find its greatest value.
