Increasing, decreasing and extrema of a function

Finding the intervals of increase, decrease and extrema of a function is both an independent task and an essential part of other tasks, in particular, full function study. Initial information about the increase, decrease and extrema of the function is given in theoretical chapter on derivative, which I highly recommend for preliminary study (or repetition)– also for the reason that the following material is based on the very essentially derivative, being a harmonious continuation of this article. Although, if time is short, then a purely formal practice of examples from today’s lesson is also possible.

And today there is a spirit of rare unanimity in the air, and I can directly feel that everyone present is burning with desire learn to explore a function using its derivative. Therefore, reasonable, kind, eternal terminology immediately appears on your monitor screens.

For what? One of the reasons is the most practical: so that it is clear what is generally required of you in a particular task!

Monotonicity of the function. Extremum points and extrema of a function

Let's consider some function. To put it simply, we assume that she continuous on the entire number line:

Just in case, let’s immediately get rid of possible illusions, especially for those readers who have recently become acquainted with intervals of constant sign of the function. Now we NOT INTERESTED, how the graph of the function is located relative to the axis (above, below, where the axis intersects). To be convincing, mentally erase the axes and leave one graph. Because that’s where the interest lies.

Function increases on an interval if for any two points of this interval connected by the relation , the inequality is true. That is, a larger value of the argument corresponds to a larger value of the function, and its graph goes “from bottom to top”. The demonstration function grows over the interval.

Likewise, the function decreases on an interval if for any two points of a given interval such that , the inequality is true. That is, a larger value of the argument corresponds to a smaller value of the function, and its graph goes “from top to bottom”. Our function decreases on intervals .

If a function increases or decreases over an interval, then it is called strictly monotonous at this interval. What is monotony? Take it literally – monotony.

You can also define non-decreasing function (relaxed condition in the first definition) and non-increasing function (softened condition in the 2nd definition). A non-decreasing or non-increasing function on an interval is called a monotonic function on a given interval (strict monotonicity is a special case of “simply” monotonicity).

The theory also considers other approaches to determining the increase/decrease of a function, including on half-intervals, segments, but in order not to pour oil-oil-oil on your head, we will agree to operate with open intervals with categorical definitions - this is clearer, and for solving many practical problems quite enough.

Thus, in my articles the wording “monotonicity of a function” will almost always be hidden intervals strict monotony(strictly increasing or strictly decreasing function).

Neighborhood of a point. Words after which students run away wherever they can and hide in horror in the corners. ...Although after the post Cauchy limits They are probably no longer hiding, but only shuddering slightly =) Don’t worry, now there will be no proofs of the theorems of mathematical analysis - I needed the surroundings to formulate the definitions more strictly extremum points. Let's remember:

Neighborhood of a point an interval that contains a given point is called, and for convenience the interval is often assumed to be symmetrical. For example, a point and its standard neighborhood:

Actually, the definitions:

The point is called strict maximum point, If exists her neighborhood, for all values ​​of which, except for the point itself, the inequality . In our specific example, this is a dot.

The point is called strict minimum point, If exists her neighborhood, for all values ​​of which, except for the point itself, the inequality . In the drawing there is point “a”.

Note : the requirement of neighborhood symmetry is not at all necessary. In addition, it is important the very fact of existence neighborhood (whether tiny or microscopic) that satisfies the specified conditions

The points are called strictly extremum points or simply extremum points functions. That is, it is a generalized term for maximum points and minimum points.

How do we understand the word “extreme”? Yes, just as directly as monotony. Extreme points of roller coasters.

As in the case of monotonicity, loose postulates exist and are even more common in theory (which, of course, the strict cases considered fall under!):

The point is called maximum point, If exists its surroundings are such that for all
The point is called minimum point, If exists its surroundings are such that for all values ​​of this neighborhood, the inequality holds.

Note that according to the last two definitions, any point of a constant function (or a “flat section” of a function) is considered both a maximum and a minimum point! The function, by the way, is both non-increasing and non-decreasing, that is, monotonic. However, we will leave these considerations to theorists, since in practice we almost always contemplate traditional “hills” and “hollows” (see drawing) with a unique “king of the hill” or “princess of the swamp”. As a variety, it occurs tip, directed up or down, for example, the minimum of the function at the point.

Oh, and speaking of royalty:
– the meaning is called maximum functions;
– the meaning is called minimum functions.

Common name - extremes functions.

Please be careful with your words!

Extremum points– these are “X” values.
Extremes– “game” meanings.

! Note : sometimes the listed terms refer to the “X-Y” points that lie directly on the GRAPH OF the function ITSELF.

How many extrema can a function have?

None, 1, 2, 3, ... etc. to infinity. For example, sine has infinitely many minima and maxima.

IMPORTANT! The term "maximum of function" not identical the term “maximum value of a function”. It is easy to notice that the value is maximum only in a local neighborhood, and there are “cooler comrades” at the top left. Likewise, “minimum of a function” is not the same as “minimum value of a function,” and in the drawing we see that the value is minimum only in a certain area. In this regard, extremum points are also called local extremum points, and the extrema – local extremes. They walk and wander nearby and global brethren. So, any parabola has at its vertex global minimum or global maximum. Further, I will not distinguish between types of extremes, and the explanation is voiced more for general educational purposes - the additional adjectives “local”/“global” should not take you by surprise.

Let’s summarize our short excursion into the theory with a test shot: what does the task “find the monotonicity intervals and extremum points of the function” mean?

The wording encourages you to find:

– intervals of increasing/decreasing function (non-decreasing, non-increasing appears much less often);

– maximum and/or minimum points (if any exist). Well, to avoid failure, it’s better to find the minimums/maximums themselves ;-)

How to determine all this? Using the derivative function!

How to find intervals of increasing, decreasing,
extremum points and extrema of the function?

Many rules, in fact, are already known and understood from lesson about the meaning of a derivative.

Tangent derivative brings the cheerful news that function is increasing throughout domain of definition.

With cotangent and its derivative the situation is exactly the opposite.

The arcsine increases over the interval - the derivative here is positive: .
When the function is defined, but not differentiable. However, at the critical point there is a right-handed derivative and a right-handed tangent, and at the other edge there are their left-handed counterparts.

I think it won’t be too difficult for you to carry out similar reasoning for the arc cosine and its derivative.

All of the above cases, many of which are tabular derivatives, I remind you, follow directly from derivative definitions.

Why explore a function using its derivative?

To better understand what the graph of this function looks like: where it goes “bottom up”, where “top down”, where it reaches minimums and maximums (if it reaches at all). Not all functions are so simple - in most cases we have no idea at all about the graph of a particular function.

It's time to move on to more meaningful examples and consider algorithm for finding intervals of monotonicity and extrema of a function:

Example 1

Find intervals of increase/decrease and extrema of the function

Solution:

1) The first step is to find domain of a function, and also take note of the break points (if they exist). In this case, the function is continuous on the entire number line, and this action is to a certain extent formal. But in a number of cases, serious passions flare up here, so let’s treat the paragraph without disdain.

2) The second point of the algorithm is due to

a necessary condition for an extremum:

If there is an extremum at a point, then either the value does not exist.

Confused by the ending? Extremum of the “modulus x” function .

The condition is necessary, but not enough, and the converse is not always true. So, it does not yet follow from the equality that the function reaches a maximum or minimum at point . A classic example has already been highlighted above - this is a cubic parabola and its critical point.

But be that as it may, the necessary condition for an extremum dictates the need to find suspicious points. To do this, find the derivative and solve the equation:

At the beginning of the first article about function graphs I told you how to quickly build a parabola using an example : “...we take the first derivative and equate it to zero: ...So, the solution to our equation: - it is at this point that the vertex of the parabola is located...”. Now, I think, everyone understands why the vertex of the parabola is located exactly at this point =) In general, we should start with a similar example here, but it is too simple (even for a teapot). In addition, there is an analogue at the very end of the lesson about derivative of a function. Therefore, let's increase the degree:

Example 2

Find intervals of monotonicity and extrema of the function

This is an example for you to solve on your own. A complete solution and an approximate final sample of the problem at the end of the lesson.

The long-awaited moment of meeting with fractional-rational functions has arrived:

Example 3

Explore a function using the first derivative

Pay attention to how variably one and the same task can be reformulated.

Solution:

1) The function suffers infinite discontinuities at points.

2) We detect critical points. Let's find the first derivative and equate it to zero:

Let's solve the equation. A fraction is zero when its numerator is zero:

Thus, we get three critical points:

3) We plot ALL detected points on the number line and interval method we define the signs of the DERIVATIVE:

I remind you that you need to take some point in the interval and calculate the value of the derivative at it and determine its sign. It’s more profitable not to even count, but to “estimate” verbally. Let's take, for example, a point belonging to the interval and perform the substitution: .

Two “pluses” and one “minus” give a “minus”, therefore, which means the derivative is negative over the entire interval.

The action, as you understand, needs to be carried out for each of the six intervals. By the way, note that the numerator factor and denominator are strictly positive for any point in any interval, which greatly simplifies the task.

So, the derivative told us that the FUNCTION ITSELF increases by and decreases by . It is convenient to connect intervals of the same type with the join icon.

At the point the function reaches its maximum:
At the point the function reaches a minimum:

Think about why you don't have to recalculate the second value ;-)

When passing through a point, the derivative does not change sign, so the function has NO EXTREMUM there - it both decreased and remained decreasing.

! Let's repeat an important point: points are not considered critical - they contain a function not determined. Accordingly, here In principle there can be no extremes(even if the derivative changes sign).

Answer: function increases by and decreases by At the point the maximum of the function is reached: , and at the point – the minimum: .

Knowledge of monotonicity intervals and extrema, coupled with established asymptotes already gives a very good idea of ​​the appearance of the function graph. A person of average training is able to verbally determine that the graph of a function has two vertical asymptotes and an oblique asymptote. Here is our hero:

Try once again to correlate the results of the study with the graph of this function.
There is no extremum at the critical point, but there is inflection point(which, as a rule, happens in similar cases).

Example 4

Find the extrema of the function

Example 5

Find monotonicity intervals, maxima and minima of the function

…it’s almost like some kind of “X in a cube” holiday today....
Soooo, who in the gallery offered to drink for this? =)

Each task has its own substantive nuances and technical subtleties, which are commented on at the end of the lesson.

As you can see, this sign of an extremum of a function requires the existence of a derivative at least to the second order at the point.

Example.

Find the extrema of the function.

Solution.

Let's start with the domain of definition:

Let's differentiate the original function:

x=1, that is, this is a point of possible extremum. We find the second derivative of the function and calculate its value at x = 1:

Therefore, by the second sufficient condition for an extremum, x=1- maximum point. Then - maximum function.

Graphic illustration.

Answer:

The third sufficient condition for the extremum of a function.

Let the function y=f(x) has derivatives up to n-th order in the -neighborhood of the point and derivatives up to n+1-th order at the point itself. Let it be.

Example.

Find the extremum points of the function .

Solution.

The original function is a rational entire function; its domain of definition is the entire set of real numbers.

Let's differentiate the function:

The derivative goes to zero at , therefore, these are points of possible extremum. Let us use the third sufficient condition for an extremum.

We find the second derivative and calculate its value at points of possible extremum (we will omit intermediate calculations):

Consequently, is the maximum point (for the third sufficient sign of extremum we have n=1 And ).

To find out the nature of the points we find the third derivative and calculate its value at these points:

Therefore, is the inflection point of the function ( n=2 And ).

It remains to deal with the point. We find the fourth derivative and calculate its value at this point:

Therefore, is the minimum point of the function.

Graphic illustration.

Answer:

The maximum point is the minimum point of the function.

10. Extrema of a function Definition of an extremum

The function y = f(x) is called increasing (decreasing) in a certain interval, if for x 1< x 2 выполняется неравенство (f(x 1) < f (x 2) (f(x 1) >f(x 2)).

If the differentiable function y = f(x) increases (decreases) on an interval, then its derivative on this interval f " (x)  0

(f " (x)  0).

Dot x O called local maximum point (minimum) function f(x), if there is a neighborhood of the point x O, for all points of which the inequality f(x) ≤ f(x о) (f(x) ≥ f(x о)) is true.

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extremes.

Extremum points

Necessary conditions for an extremum. If the point x O is an extremum point of the function f(x), then either f " (x o) = 0, or f (x o) does not exist. Such points are called critical, and the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let x O- critical point. If f "(x) when passing through a point x O changes the plus sign to minus, then at the point x O the function has a maximum, otherwise it has a minimum. If, when passing through the critical point, the derivative does not change sign, then at the point x O there is no extreme.

Second sufficient condition. Let the function f(x) have a derivative f " (x) in the vicinity of the point x O and the second derivative at the point itself x O. If f "(x o) = 0, >0 (<0), то точка x O is the local minimum (maximum) point of the function f(x). If =0, then you need to either use the first sufficient condition or use higher derivatives.

On a segment, the function y = f(x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) = 6x 2 - 30x +36 = 6(x ​​-2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. So as when passing through the point x 1 = 2 the derivative changes its sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3 the derivative changes its sign from minus to plus, therefore at the point x 2 = 3 the function has a minimum. Having calculated the values ​​of the function at the points x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.

The extremum point of a function is the point in the domain of definition of the function at which the value of the function takes on a minimum or maximum value. The values ​​of the function at these points are called extrema (minimum and maximum) of the function.

Definition. Dot x1 function domain f(x) is called maximum point of the function , if the value of the function at this point is greater than the values ​​of the function at points sufficiently close to it, located to the right and left of it (that is, the inequality holds f(x0 ) > f(x 0 + Δ x) x1 maximum.

Definition. Dot x2 function domain f(x) is called minimum point of the function, if the value of the function at this point is less than the values ​​of the function at points sufficiently close to it, located to the right and left of it (that is, the inequality holds f(x0 ) < f(x 0 + Δ x) ). In this case we say that the function has at the point x2 minimum.

Let's say point x1 - maximum point of the function f(x) . Then in the interval up to x1 function increases, therefore the derivative of the function is greater than zero ( f "(x) > 0 ), and in the interval after x1 the function decreases, therefore, derivative of a function less than zero ( f "(x) < 0 ). Тогда в точке x1

Let us also assume that the point x2 - minimum point of the function f(x) . Then in the interval up to x2 the function is decreasing, and the derivative of the function is less than zero ( f "(x) < 0 ), а в интервале после x2 the function is increasing, and the derivative of the function is greater than zero ( f "(x) > 0 ). In this case also at the point x2 the derivative of the function is zero or does not exist.

Fermat's theorem (a necessary sign of the existence of an extremum of a function). If the point x0 - extremum point of the function f(x) then at this point the derivative of the function is equal to zero ( f "(x) = 0 ) or does not exist.

Definition. The points at which the derivative of a function is zero or does not exist are called critical points .

Example 1. Let's consider the function.

At the point x= 0 the derivative of the function is zero, therefore the point x= 0 is the critical point. However, as can be seen on the graph of the function, it increases throughout the entire domain of definition, so the point x= 0 is not the extremum point of this function.

Thus, the conditions that the derivative of a function at a point is equal to zero or does not exist are necessary conditions for an extremum, but not sufficient, since other examples of functions can be given for which these conditions are met, but the function does not have an extremum at the corresponding point. That's why there must be sufficient evidence, allowing one to judge whether there is an extremum at a particular critical point and what kind of extremum it is - maximum or minimum.

Theorem (the first sufficient sign of the existence of an extremum of a function). Critical point x0 f(x) if, when passing through this point, the derivative of the function changes sign, and if the sign changes from “plus” to “minus”, then it is a maximum point, and if from “minus” to “plus”, then it is a minimum point.

If near the point x0 , to the left and to the right of it, the derivative retains its sign, this means that the function either only decreases or only increases in a certain neighborhood of the point x0 . In this case, at the point x0 there is no extreme.

So, to determine the extremum points of the function, you need to do the following :

1. Find the derivative of the function.
2. Equate the derivative to zero and determine the critical points.
3. Mentally or on paper, mark the critical points on the number line and determine the signs of the derivative of the function in the resulting intervals. If the sign of the derivative changes from “plus” to “minus”, then the critical point is the maximum point, and if from “minus” to “plus”, then the minimum point.
4. Calculate the value of the function at the extremum points.

Example 2. Find the extrema of the function .

Solution. Let's find the derivative of the function:

Let's equate the derivative to zero to find the critical points:

.

Since for any values ​​of “x” the denominator is not equal to zero, we equate the numerator to zero:

Got one critical point x= 3 . Let us determine the sign of the derivative in the intervals delimited by this point:

in the interval from minus infinity to 3 - a minus sign, that is, the function decreases,

in the interval from 3 to plus infinity there is a plus sign, that is, the function increases.

That is, period x= 3 is the minimum point.

Let's find the value of the function at the minimum point:

Thus, the extremum point of the function is found: (3; 0), and it is the minimum point.

Theorem (the second sufficient sign of the existence of an extremum of a function). Critical point x0 is the extremum point of the function f(x) if the second derivative of the function at this point is not equal to zero ( f ""(x) ≠ 0 ), and if the second derivative is greater than zero ( f ""(x) > 0 ), then the maximum point, and if the second derivative is less than zero ( f ""(x) < 0 ), то точкой минимума.

Note 1. If at the point x0 If both the first and second derivatives vanish, then at this point it is impossible to judge the presence of an extremum based on the second sufficient criterion. In this case, you need to use the first sufficient criterion for the extremum of a function.

Remark 2. The second sufficient criterion for the extremum of a function is not applicable even when the first derivative does not exist at a stationary point (then the second derivative does not exist either). In this case, you also need to use the first sufficient sign of an extremum of a function.

Local nature of the extrema of the function

From the above definitions it follows that the extremum of a function is local in nature - it is the largest and smallest value of the function compared to nearby values.

Let's say you're looking at your earnings over a period of one year. If in May you earned 45,000 rubles, and in April 42,000 rubles and in June 39,000 rubles, then May earnings are the maximum of the earnings function compared to nearby values. But in October you earned 71,000 rubles, in September 75,000 rubles, and in November 74,000 rubles, so October earnings are the minimum of the earnings function compared to nearby values. And you can easily see that the maximum among the values ​​of April-May-June is less than the minimum of September-October-November.

Generally speaking, on an interval a function can have several extrema, and it may turn out that some minimum of the function is greater than any maximum. So, for the function shown in the figure above, .

That is, one should not think that the maximum and minimum of a function are, respectively, its largest and smallest values ​​on the entire segment under consideration. At the maximum point, the function has the greatest value only in comparison with those values ​​​​that it has at all points sufficiently close to the maximum point, and at the minimum point it has the smallest value only in comparison with those values ​​​​that it has at all points sufficiently close to the minimum point.

Therefore, we can clarify the above concept of extremum points of a function and call minimum points local minimum points, and maximum points local maximum points.

We look for the extrema of the function together

Example 3.

Solution: The function is defined and continuous on the entire number line. Its derivative also exists on the entire number line. Therefore, in this case, the critical points are only those at which, i.e. , from where and . Critical points and divide the entire domain of definition of the function into three intervals of monotonicity: . Let's select one control point in each of them and find the sign of the derivative at this point.

For the interval, the control point can be: find. Taking a point in the interval, we get, and taking a point in the interval, we have. So, in the intervals and , and in the interval . According to the first sufficient criterion for an extremum, there is no extremum at the point (since the derivative retains its sign in the interval), and at the point the function has a minimum (since the derivative changes sign from minus to plus when passing through this point). Let's find the corresponding values ​​of the function: , a . In the interval the function decreases, since in this interval , and in the interval it increases, since in this interval .

To clarify the construction of the graph, we find the points of intersection of it with the coordinate axes. When we obtain an equation whose roots are and , i.e., two points (0; 0) and (4; 0) of the graph of the function are found. Using all the information received, we build a graph (see the beginning of the example).

Example 4. Find the extrema of the function and build its graph.

The domain of definition of a function is the entire number line, except for the point, i.e. .

To shorten the study, you can use the fact that this function is even, since . Therefore, its graph is symmetrical about the axis Oy and the study can only be performed for the interval.

Finding the derivative and critical points of the function:

1) ;

2) ,

but the function suffers a discontinuity at this point, so it cannot be an extremum point.

Thus, the given function has two critical points: and . Taking into account the parity of the function, we will check only the point using the second sufficient criterion for an extremum. To do this, we find the second derivative and determine its sign at: we get . Since and , it is the minimum point of the function, and .

To get a more complete picture of the graph of a function, let’s find out its behavior at the boundaries of the domain of definition:

(here the symbol indicates the desire x to zero from the right, and x remains positive; similarly means aspiration x to zero from the left, and x remains negative). Thus, if , then . Next, we find

,

those. if , then .

The graph of a function has no intersection points with the axes. The picture is at the beginning of the example.

We continue to search for extrema of the function together

Example 8. Find the extrema of the function.

Solution. Let's find the domain of definition of the function. Since the inequality must be satisfied, we obtain from .

Let's find the first derivative of the function:

Let's find the critical points of the function.

>>Extrema

Extremum of the function

Definition of extremum

Function y = f(x) is called increasing (decreasing) in a certain interval, if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f (x 2)).

If the differentiable function y = f (x) increases (decreases) on an interval, then its derivative on this interval f " (x)> 0

(f"(x)< 0).

Dot x O called local maximum point (minimum) function f (x) if there is a neighborhood of the point x o, for all points of which the inequality f (x) is true≤ f (x o ) (f (x )≥ f (x o )).

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extremes.

Extremum points

Necessary conditions for an extremum . If the point x O is the extremum point of the function f (x), then either f " (x o ) = 0, or f(x o ) does not exist. Such points are called critical, and the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let x O - critical point. If f" (x ) when passing through a point x O changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If, when passing through the critical point, the derivative does not change sign, then at the point x O there is no extreme.

Second sufficient condition. Let the function f(x) have
f" (x ) in the vicinity of the point x O and the second derivative at the point itself x o. If f"(x o) = 0, >0 ( <0), то точка x o is the local minimum (maximum) point of the function f (x). If =0, then you need to either use the first sufficient condition or involve higher ones.

On a segment, the function y = f (x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22.

Solution. Because f " (

Problems of finding the extremum of a function

Example 3.23. a

Solution. x And y y
0 ≤ x≤
> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions kv. units).

Example 3.24. p ≈

Solution. p p
S"
R = 2, H = 16/4 = 4.

Example 3.22.Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Because f " (x ) = 6x 2 - 30x +36 = 6(x ​​-2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. Since when passing through the point x 1 = 2 the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3, the derivative changes its sign from minus to plus, so at the point x 2 = 3 the function has a minimum. Having calculated the function values ​​at the points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23.It is necessary to build a rectangular area near the stone wall so that it is fenced off on three sides with wire mesh, and the fourth side is adjacent to the wall. For this there is a linear meters of mesh. At what aspect ratio will the site have the largest area?

Solution.Let us denote the sides of the platform by x And y. The area of ​​the site is S = xy. Let y- this is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must be satisfied. Therefore y = a - 2x and S = x (a - 2x), where
0 ≤ x≤ a /2 (the length and width of the area cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2 × a/4 =a/2. Because the x = a /4 is the only critical point; let’s check whether the sign of the derivative changes when passing through this point. At x a /4 S "> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions S(a/4) = a/4(a - a/2) = a 2 /8 (kv. units). Since S is continuous on and its values ​​at the ends S(0) and S(a /2) are equal to zero, then the value found will be the largest value of the function. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24.It is required to manufacture a closed cylindrical tank with a capacity of V=16 p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used for its manufacture?

Solution.The total surface area of ​​the cylinder is S = 2 p R(R+H). We know the volume of the cylinder V = p R 2 Н Þ Н = V/ p R 2 =16 p / p R2 = 16/R2. So S(R) = 2 p (R 2 +16/R). We find the derivative of this function:
S"(R) = 2 p (2R- 16/R 2) = 4 p (R- 8/R 2). S" (R) = 0 at R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

Lesson on the topic: "Finding extrema points of functions. Examples"

Additional materials
Dear users, do not forget to leave your comments, reviews, wishes! All materials have been checked by an anti-virus program.

Manuals and simulators in the Integral online store for grade 10 from 1C
We solve problems in geometry. Interactive construction tasks for grades 7-10
Software environment "1C: Mathematical Constructor 6.1"

What we will study:
1. Introduction.
2. Minimum and maximum points.

4. How to calculate extrema?
5. Examples.

Introduction to extrema of functions

Guys, let's look at the graph of some function:

Notice that the behavior of our function y=f (x) is largely determined by two points x1 and x2. Let's take a closer look at the graph of the function at and around these points. Up to point x2 the function increases, at point x2 there is an inflection, and immediately after this point the function decreases to point x1. At point x1 the function bends again, and after that it increases again. For now, we will call points x1 and x2 inflection points. Let's draw tangents at these points:

The tangents at our points are parallel to the x-axis, which means that the slope of the tangent is zero. This means that the derivative of our function at these points is equal to zero.

Let's look at the graph of this function:

It is impossible to draw tangent lines at points x2 and x1. This means that the derivative does not exist at these points. Now let's look again at our points on the two graphs. Point x2 is the point at which the function reaches its greatest value in some region (near point x2). Point x1 is the point at which the function reaches its smallest value in some region (near point x1).

Minimum and maximum points

Definition: The point x= x0 is called the minimum point of the function y=f(x) if there is a neighborhood of the point x0 in which the inequality holds: f(x) ≥ f(x0).

Definition: The point x=x0 is called the maximum point of the function y=f(x) if there is a neighborhood of the point x0 in which the inequality holds: f(x) ≤ f(x0).

Guys, what is a neighborhood?

Definition: A neighborhood of a point is a set of points containing our point and those close to it.

We can set the neighborhood ourselves. For example, for a point x=2, we can define a neighborhood in the form of points 1 and 3.

Let's return to our graphs, look at point x2, it is larger than all other points from a certain neighborhood, then, by definition, this is the maximum point. Now let's look at point x1, it is smaller than all other points from a certain neighborhood, then by definition it is a minimum point.

Guys, let's introduce the notation:

Y min - minimum point,
y max - maximum point.

Important! Guys, do not confuse the maximum and minimum points with the smallest and largest value of the function. The minimum and maximum values ​​are sought over the entire domain of definition of a given function, and the minimum and maximum points are sought in a certain neighborhood.

Extrema of the function

For minimum and maximum points there is a common term - extremum points.

Extremum (lat. extremum – extreme) – the maximum or minimum value of a function on a given set. The point at which the extremum is reached is called the extremum point.

Accordingly, if a minimum is reached, the extremum point is called a minimum point, and if a maximum is reached, it is called a maximum point.

How to look for extrema of a function?

Let's go back to our charts. At our points, the derivative either vanishes (on the first graph) or does not exist (on the second graph).

Then we can make an important statement: If the function y= f(x) has an extremum at the point x=x0, then at this point the derivative of the function is either zero or does not exist.

The points at which the derivative is equal to zero are called stationary.

The points at which the derivative of a function does not exist are called critical.

How to calculate extremes?

Guys, let's go back to the first graph of the function:

Analyzing this graph, we said: up to point x2 the function increases, at point x2 an inflection occurs, and after this point the function decreases to point x1. At point x1 the function bends again, and after that the function increases again.

Based on such reasoning, we can conclude that the function at extremum points changes the nature of monotonicity, and therefore the derivative function changes sign. Recall: if a function decreases, then the derivative is less than or equal to zero, and if the function increases, then the derivative is greater than or equal to zero.

Let us summarize the knowledge gained with the following statement:

Theorem: A sufficient condition for an extremum: let the function y=f(x) be continuous on some interval X and have a stationary or critical point x= x0 inside the interval. Then:

• If this point has a neighborhood in which f’(x)>0 holds for x x0, then point x0 is the minimum point of the function y= f(x).
• If this point has a neighborhood in which f'(x) holds for x 0 and x> x0. If this point has a neighborhood in which both to the left and to the right of the point x0 the signs of the derivative are the same, then at the point x0 there is no extreme.

To solve problems, remember these rules: If the signs of derivatives are defined then:

Algorithm for studying a continuous function y= f(x) for monotonicity and extrema:

• Find the derivative of y'.
• Find stationary points (the derivative is zero) and critical points (the derivative does not exist).
• Mark stationary and critical points on the number line and determine the signs of the derivative on the resulting intervals.
• Based on the above statements, draw a conclusion about the nature of the extremum points.

Examples of finding extreme points

1) Find the extremum points of the function and determine their nature: y= 7+ 12*x - x 3

Solution: Our function is continuous, then we will use our algorithm:
a) y"= 12 - 3x 2,
b) y"= 0, at x= ±2,

Point x= -2 is the minimum point of the function, point x= 2 is the maximum point of the function.
Answer: x= -2 is the minimum point of the function, x= 2 is the maximum point of the function.

2) Find the extremum points of the function and determine their nature.

3) Find the extremum points of the function y= x - 2cos(x) and determine their nature, for -π ≤ x ≤ π.

Solution: Our function is continuous, let's use our algorithm:
a) y"= 1 + 2sin(x),
b) find the values ​​in which the derivative is equal to zero: 1 + 2sin(x)= 0, sin(x)= -1/2,
because -π ≤ x ≤ π, then: x= -π/6, -5π/6,
c) mark stationary points on the number line and determine the signs of the derivative: d) look at our figure, which shows the rules for determining extrema.
Point x= -5π/6 is the maximum point of the function.
Point x= -π/6 is the minimum point of the function.
Answer: x= -5π/6 is the maximum point of the function, x= -π/6 is the minimum point of the function.

4) Find the extremum points of the function and determine their nature:

Problems to solve independently