How to find the greatest value of the function without graphics. The greatest and smallest values \u200b\u200bof the function on the segment

The greatest (smallest) function of the function is the largest (small) adopted order value on the interval under consideration.

To find the greatest or smallest function of the function you need:

1. Check which stationary points are included in the specified segment.
2. Calculate the value of the function at the ends of the segment and in stationary points from clause 3
3. Select the most or smallest value from the results obtained.

To find a maximum or minimum point you need:

1. Find the derivative function $ f "(x) $
2. Find stationary points, deciding the equation $ F "(x) \u003d 0 $
3. Dispatch the derivative of the functions on multipliers.
4. Hold the coordinate direct, place stationary points on it and determine the signs of the derivative in the obtained intervals, using the record of clause 3.
5. Find a maximum point or a minimum of rule: if at the point the derivative changes the sign from the plus to minus, then it will be a maximum point (if with a minus on plus, then it will be a minimum point). In practice, it is convenient to use the image of the arrows on the intervals: on the interval, where the derivative is positive, the arrow is drawn up and vice versa.

Table of derivatives of some elementary functions:

Basic Differentiation Rules

1. The derivative of the amount and difference is equal to the derivative of each term

$ (f (x) ± g (x)) '\u003d f' (x) ± g '(x) $

Find the derivative function $ f (x) \u003d 3x ^ 5 - Cosx + (1) / (x) $

The derivative of the amount and difference is equal to the derivative of each terms

$ f '(x) \u003d (3x ^ 5)' - (cosx) '+ ((1) / (x)) "\u003d 15x ^ 4 + sinx- (1) / (x ^ 2) $

2. Derivative work.

$ (f (x) ∙ g (x)) '\u003d f' (x) ∙ g (x) + f (x) ∙ g (x) '$

Find a derivative $ F (x) \u003d 4x ∙ COSX $

$ F '(x) \u003d (4x)' ∙ COSX + 4X ∙ (COSX) '\u003d 4 ∙ COSX-4X ∙ SINX $

3. Private derivative

$ ((F (x)) / (G (x))) "\u003d (f ^" (x) ∙ g (x) -f (x) ∙ g (x) ") / (g ^ 2 (x) ) $

Find the derivative $ F (x) \u003d (5x ^ 5) / (E ^ x) $

$ f "(x) \u003d ((5x ^ 5)" ∙ E ^ x-5x ^ 5 ∙ (E ^ x) ") / ((E ^ x) ^ 2) \u003d (25x ^ 4 ∙ E ^ X- 5x ^ 5 ∙ E ^ x) / ((E ^ x) ^ 2) $

4. The derivative of the complex function is equal to the product of the derivative of the external function on the derivative of the internal function

$ F (G (x)) '\u003d F' (G (x)) ∙ g '(x) $

$ f '(x) \u003d cos' (5x) ∙ (5x)' \u003d - sin (5x) ∙ 5 \u003d -5sin (5x) $

Find the point of the minimum function $ y \u003d 2x-ln\u2061 (x + 11) + $ 4

1. Find ... Functions: $ x + 11\u003e 0; x\u003e -11 $

2. Find the derivative function $ y "\u003d 2- (1) / (x + 11) \u003d (2x + 22-1) / (x + 11) \u003d (2x + 21) / (x + 11) $ 2.

3. Find stationary points, equating the derivative to zero

$ (2x + 21) / (x + 11) \u003d 0 $

The fraction is zero if the numerator is zero, and the denominator is not zero

$ 2x + 21 \u003d 0; x ≠ -11 $

4. Draw the coordinate direct, lay stationary points on it and determine the signs of the derivative in the intervals obtained. To do this, we will substitute in a derivative any number of the extreme right area, for example, zero.

$ y "(0) \u003d (2 ∙ 0 + 21) / (0 + 11) \u003d (21) / (11)\u003e 0 $

5. At a minimum point, the derivative changes a sign from a minus on plus, therefore, a point of $ -10.5 $ is a minimum point.

Answer: $ -10.5 $

Find the greatest value of the function $ y \u003d 6x ^ 5-90x ^ 3-5 $ on the segment $ [- 5; 1] $

1. Find the derivative of the function $ y '\u003d 30x ^ 4-270x ^ 2 $

2. Ensure derivatives to zero and find stationary points

$ 30x ^ 4-270x ^ 2 \u003d 0 $

I will carry out a total multiplier $ 30x ^ $ 2 for brackets

$ 30x ^ 2 (x ^ 2-9) \u003d $ 0

$ 30x ^ 2 (x-3) (x + 3) \u003d 0 $

We equate every multiplier to zero

$ x ^ 2 \u003d 0; x-3 \u003d 0; x + 3 \u003d 0 $

$ x \u003d 0; x \u003d 3; x \u003d -3 $

3. Select stationary points that belong to a given segment $ [- 5; 1] $

We are suitable for stationary points $ x \u003d 0 $ and $ x \u003d -3 $

4. Calculate the value of the function at the ends of the segment and in stationary points from clause 3

What is the extremum function and what is the necessary extremma condition?

Extreme function is called maximum and minimum function.

The prerequisite of the maximum and minimum (extremum) function is as follows: if the function f (x) has an extremum at point x \u003d 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 \u003d or can contact zero, in infinity or not to exist without the function to have an extremum at this point.

What is the sufficient condition of the extremum function (maximum or minimum)?

First condition:

If in sufficient proximity to point x \u003d a derivative f? (X) is positive to the left of a and negative to the right of a, then at the point itself x \u003d and the function f (x) has maximum

If in sufficient proximity to the point x \u003d and the derivative f? (X) is negative from the left of the A and positive to the right of the A, then at the point itself x \u003d and the function f (x) has minimum Provided that the function f (x) is continuous here.

Instead, you can use the second sufficient condition for the extremum function:

Let at point x \u003d a first derivative f? (X) refers to zero; If the second derivative f ?? (a) is negative, then the function f (x) has at the point x \u003d a maximum, if a positive is minimum.

What is a critical point function and how to find it?

This is the value of the function argument, in which the function has an extremum (i.e. maximum or minimum). To find it, you need find a derivative Functions f? (x) and equating it to zero, solve equation f? (x) \u003d 0. The roots of this equation, as well as those points in which there is no derivative of this function are critical points, i.e. the values \u200b\u200bof the argument at which the extremum may be. They can be easily defined by looking at derivative graph: We are interested in those values \u200b\u200bof the argument, in which the graph of the function crosses the abscissa axis (OH axis) and those in which the graphs tolerate breaks.

For example, find extreme parabolla.

Function y (x) \u003d 3x2 + 2x - 50.

Derived function: y? (X) \u003d 6x + 2

We solve the equation: y? (X) \u003d 0

6x + 2 \u003d 0, 6x \u003d -2, x \u003d -2 / 6 \u003d -1/3

In this case, the critical point is x0 \u003d -1 / 3. It is with the meaning of the argument that the function has extremum. So that to find, We substitute an expression for a function instead of "x" found number:

y0 \u003d 3 * (- 1/3) 2 + 2 * (- 1/3) - 50 \u003d 3 * 1/9 - 2/3 - 50 \u003d 1/3 - 2/3 - 50 \u003d -1/3 - 50 \u003d -50,333.

How to determine the maximum and minimum of the function, i.e. Her largest and smallest meanings?

If the sign of the derivative during the transition through the critical point x0 is changing from the "plus" to "minus", then x0 is maximum point; if the sign of the derivative changes with a minus on plus, then x0 is point of minimum; If the sign does not change, then at the point x0, no maximum, no minimum.

For the considered example:

We take an arbitrary value of the argument to the left of the critical point: x \u003d -1

At x \u003d -1, the value of the derivative would be? (- 1) \u003d 6 * (- 1) + 2 \u003d -6 + 2 \u003d -4 (that is, the sign is "minus").

Now take an arbitrary value of the argument to the right of the critical point: x \u003d 1

At x \u003d 1, the value of the derivative will be (1) \u003d 6 * 1 + 2 \u003d 6 + 2 \u003d 8 (i.e. the sign is "plus").

As we see, the derivative during the transition through the critical point changed the sign with a minus on the plus. So, with a critical value x0, we have a minimum point.

The greatest and smallest value of the function at the interval (on the segment) are found along the same procedure, only taking into account the fact that, perhaps, not all critical points will lie inside the specified interval. Those critical points that are for the range of intervals must be excluded from consideration. If only one critical point is inside the interval - it will be either maximum or a minimum. In this case, to determine the greatest and smallest function values, we also take into account the values \u200b\u200bof the function at the ends of the interval.

For example, find the greatest and smallest values \u200b\u200bof the function.

y (x) \u003d 3sin (x) - 0,5x

at intervals:

So, derived function -

y? (x) \u003d 3cos (x) - 0.5

We solve the equation 3cos (x) - 0.5 \u003d 0

cos (x) \u003d 0.5 / 3 \u003d 0,16667

x \u003d ± Arccos (0.16667) + 2πk.

We find critical points at the interval [-9; nine]:

x \u003d Arccos (0.16667) - 2π * 2 \u003d -11,163 (not included in the interval)

x \u003d -arccos (0.16667) - 2π * 1 \u003d -7,687

x \u003d Arccos (0.16667) - 2π * 1 \u003d -4,88

x \u003d -arccos (0.16667) + 2π * 0 \u003d -1,403

x \u003d Arccos (0.16667) + 2π * 0 \u003d 1.403

x \u003d -arccos (0.16667) + 2π * 1 \u003d 4.88

x \u003d Arccos (0.16667) + 2π * 1 \u003d 7,687

x \u003d -arccos (0.16667) + 2π * 2 \u003d 11,163 (not included in the interval)

We find the values \u200b\u200bof the function at critical values \u200b\u200bof the argument:

y (-7,687) \u003d 3cos (-7,687) - 0.5 \u003d 0,885

y (-4.88) \u003d 3cos (-4,88) - 0.5 \u003d 5,398

y (-1,403) \u003d 3cos (-1,403) - 0.5 \u003d -2,256

y (1.403) \u003d 3cos (1.403) - 0.5 \u003d 2,256

y (4,88) \u003d 3cos (4,88) - 0.5 \u003d -5,398

y (7,687) \u003d 3COS (7,687) - 0.5 \u003d -0,885

It can be seen that on the interval [-9; 9] The greatest value of the function has at x \u003d -4.88:

x \u003d -4.88, y \u003d 5,398,

and the smallest - at x \u003d 4.88:

x \u003d 4.88, y \u003d -5,398.

On the interval [-6; -3] We have only one critical point: x \u003d -4.88. The value of the function at x \u003d -4.88 is equal to y \u003d 5,398.

We find the value of the function at the ends of the interval:

y (-6) \u003d 3cos (-6) - 0.5 \u003d 3,838

y (-3) \u003d 3cos (-3) - 0.5 \u003d 1,077

On the interval [-6; -3] have the greatest value of the function

y \u003d 5,398 at x \u003d -4.88

the smallest value is

y \u003d 1,077 at x \u003d -3

How to find points inflection graphics function and determine the parties of bulge and concave?

To find all the blinking points of the line y \u003d f (x), it is necessary to find the second derivative, to equate it to zero (solve the equation) and experience all those values \u200b\u200bx for which the second derivative is zero, infinite or does not exist. If during the transition through one of these values, the second derivative changes the sign, then the function graph has at this point. If it does not change, then the inflection is not.

Roots equation f? (x) \u003d 0, as well as possible points of breaking of the function and the second derivative divide the area of \u200b\u200bdetermining the function to a number of intervals. The bulge at each of their intervals is determined by the sign of the second derivative. If the second derivative at the point on the interval under study is positive, then the line y \u003d f (x) is facing here concave upwards, and if negative is the book.

How to find extremums of two variables?

To find the extremmas function f (x, y), differentiated in the area of \u200b\u200bits task, you need:

1) find critical points, and for this - solve the system of equations

fx? (x, y) \u003d 0, fu? (x, y) \u003d 0

2) for each critical point P0 (a; b) to explore whether the difference sign remains unchanged

for all points (x; y), close to P0. If the difference retains a positive sign, then at the point P0 we have a minimum, if negative is the maximum. If the difference does not save the sign, then there is no extremum at P0.

Similarly, the extremums of the function with a greater number of arguments are determined.

The study of such an object of mathematical analysis as a function has a large value And in other areas of science. For example, in economic analysis is constantly required to evaluate the behavior. functions Profit, namely to determine its largest value and develop a strategy for its achievement.

Instruction

Study of behavior Any should always be started with the search for the definition area. Usually, by a specific task, it is required to determine the largest value functions either on the whole area, or on its specific interval with open or closed borders.

Based on the greatest value functions y (x0), in which for any point of the definition area, the inequality y (x0) ≥ y (x) is performed (x ≠ x0). Graphically, this point will be highest, if you arrange the values \u200b\u200bof the argument along the abscissa axis, and the function itself along the ordinate axis.

To determine the greatest value functionsFollow the algorithm out of three stages. Note that you should be able to work with one-sided and, as well as calculate the derivative. So, let some function y (x) be asked and it is required to find its largest value At some interval with boundary values \u200b\u200bA and V.

Find out whether this interval is in the field of definition functions. To do this, it is necessary to find it, considering all possible limitations: the presence in the expression of the fraction, square root, etc. The definition area is a plurality of argument values \u200b\u200bin which the function makes sense. Determine whether this interval is a subset. If so, then go to the next stage.

Find a derivative functions And solve the obtained equation, equating the derivative to zero. Thus, you will get the values \u200b\u200bof the so-called stationary points. Rate whether at least one of them interval A, V.

Consider these points on the third stage, substitute their values \u200b\u200bto the function. Depending on the type of interval, make the following additional steps. If there is a segment of the form [A, B], the boundary points are included in the interval, they say brackets. Calculate values functions at x \u003d a and x \u003d V. If an open interval (A, B), boundary values \u200b\u200bare painted, i.e. Do not enter it. Decide unilateral limits for X → A and X → B. The combined interval of the form [A, B) or (A, B], one of the boundaries of which belongs to him, the other - no. Find a one-sided limit at x, seeking to have a paint value, and then substitute an infinite two-sided interval (-∞, + ∞) or unilateral infinite intervals of the form:, (-∞, b). For valid limits, A and in act according to the already described principles, and for infinite, look for the limits for X → -∞ and x → + ∞, respectively.

Task at this stage

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

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

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

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


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

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

Example 1.

In a limited closed area

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

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

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

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

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

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

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

(Ii) Explore the border of the region.

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

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

Alternatively, you can arrange and so:

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

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

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

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

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

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

The geometric situation is related to the previous item:

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

We explore the second end of the segment:

Using a function , Perform a check check:

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

Ends of cut already investigated, but on the draft still check, whether we have found the function correctly :
- coincided with the result of the 1st subparagraph;
- coincided with the result of the 2nd subparagraph.

It remains to find out if there is something interesting inside the segment:

- there is! Substituting the line to the equation, we get the ordinate of this "Interest":

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

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

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

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

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

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

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

Example 2.

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

Example 3.

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

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

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

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

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

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

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

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

Example 4.

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

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

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

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

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

I) find stationary points:

Idiot's dream system :)

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

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

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

1) if, then

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

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

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

Control:

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

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


Calculate the functions of the function at found points:

Specify the function yourself.

Now carefully study the conquered trophies and write down answer:

These are "candidates", so "candidates"!

For self solutions:

Example 5.

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

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

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

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

Solutions and answers:

Example 2: Decision: Show area in the drawing:

In practice, it is quite often necessary to use a derivative in order to calculate the largest and smallest function value. We carry out this action when we find out how to minimize costs, increase profits, calculate the optimal load on production, etc., that is, in cases where you need to determine the optimal value of any parameter. To solve such tasks correctly, it is necessary to understand well what the greatest and smallest value of the function.

Yandex.rtb R-A-339285-1

Usually we define these values \u200b\u200bwithin a certain interval X, which may in turn correspond to the entire field of defining the function or part of it. This may be like a segment [a; b] and an open interval (a; b), (a; b], [a; b), an infinite interval (a; b), (a; b], [a; b) or an infinite gap - ∞; a, (- ∞; a], [a; + ∞), (- ∞; + ∞).

In this material, we will describe how the largest and smallest value of an explicitly specified function with one variable y \u003d f (x) y \u003d f (x) is calculated.

Main definitions

Let's start, as always, with the formulation of basic definitions.

Definition 1.

The greatest value of the function y \u003d f (x) at some gap is maxy \u003d f (x 0) x ∈ X, which, with any meaning xx ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x 0).

Definition 2.

The smallest value of the function y \u003d f (x) at some gap is a MINX ∈ X Y \u003d F (x 0), which, with any value x ∈ X, x ≠ x 0 makes the inequality F (x F (x) ≥ f (x 0).

These definitions are quite obvious. It is even easier to say so: the greatest value of the function is its greatest value at the known interval at the abscissa x 0, and the smallest is the smallest value at the same interval at x 0.

Definition 3.

Stationary points are the values \u200b\u200bof the function argument, in which its derivative is referred to 0.

Why do we need to know what landfill points? To answer this question, you need to remember the farm theorem. It follows that a stationary point is such a point in which the extremum of the differentiable function is located (that is, its local minimum or maximum). Consequently, the function will take the smallest or most important at some interval in one of the stationary points.

Another function can take the largest or smallest value at those points in which the function itself is defined, and its first derivative does not exist.

The first question that occurs when studying this topic: In all cases, can we determine the greatest or smallest value of the function on a given segment? No, we cannot do this when the boundaries of the specified gap will coincide with the boundaries of the definition area, or if we are dealing with an infinite interval. It also happens that the function in a given segment or at infinity will be infinitely small or infinitely large values. In these cases, it is not possible to determine the greatest and / or the smallest value.

More understandable these moments will be after the image on the schedules:

The first drawing shows us a function that takes the greatest and smallest values \u200b\u200b(M a x y and m i n y) in stationary points located on the segment [- 6; 6].

We analyze in detail the case specified on the second chart. Change the segment value to [1; 6] and we obtain that the greatest value of the function will be achieved at the point with the abscissa on the right boundary of the interval, and the smallest - in the stationary point.

In the third drawing of the abscissa, points are the boundary points of the segment [- 3; 2]. They correspond to the greatest and smallest value of the specified function.

Now look at the fourth drawing. In it, the function takes M a x Y (the largest value) and M i n y (the smallest value) in the stationary points in the open interval (- 6; 6).

If we take the interval [1; 6), it can be said that the smallest value of the function on it will be achieved in a stationary point. It will be unknown as the greatest value. The function could take the highest value at x, equal to 6, if x \u003d 6 belonged to the interval. This case is drawn on the chart 5.

On Graph 6, the smallest value of this function acquires on the right boundary of the interval (- 3; 2], and we cannot make certain conclusions about the greatest value.

In Figure 7, we see that the function will have M a x Y in a stationary point having an abscissa equal to 1. The smallest function will reach on the border of the interval on the right side. On the minus infinity, the values \u200b\u200bof the function will be asymptotically approaching to y \u003d 3.

If we take the interval x ∈ 2; + ∞, we will see that the specified function will not take on it the smallest or the greatest value. If X strives for 2, the values \u200b\u200bof the function will strive for minus infinity, since the straight line X \u003d 2 is a vertical asymptota. If the abscissa tends to plus infinity, then the values \u200b\u200bof the function will be asymptotically approached y \u003d 3. This case is shown in Figure 8.

At this point, we present a sequence of actions that need to be performed for finding the largest or smallest value of the function on some segment.

1. To begin with, we find the field definition area. Check whether it is contained in the condition of the segment.
2. Now we calculate the points contained in this segment, in which there is no first derivative. Most often, they can be found in the functions whose argument is recorded under the sign of the module, or in power functions, whose indicator is a fractional rational number.
3. Next, find out which stationary points will fall into a given segment. To do this, it is necessary to calculate the function derivative, then equate it to 0 and solve the equation resulting in the end, after which it is possible to choose the appropriate roots. If we do not succeed in a single stationary point or they will not fall into a given segment, then we go to the next step.
4. We define what values \u200b\u200bwill receive the function in the specified stationary points (if any), or at those points in which there is no first derivative (if any), or calculate the values \u200b\u200bfor x \u003d a and x \u003d b.
5. 5. We turned out a number of functions of the function, of which you now need to choose the most and the smallest. This will be the greatest and smallest values \u200b\u200bof the functions that we need to find.

Let's see how to properly apply this algorithm when solving tasks.

Example 1.

Condition: The function y \u003d x 3 + 4 x 2 is specified. Determine its largest and smallest value on segments [1; 4] and [- 4; - one ] .

Decision:

Let's start with the location of the definition area of \u200b\u200bthis function. In this case, it will have many valid numbers, except 0. In other words, D (y): x ∈ (- ∞; 0) ∪ 0; + ∞. Both segments specified in the condition will be within the definition area.

Now calculate the derivative function according to the Differentiation Ruings:

y "\u003d x 3 + 4 x 2" \u003d x 3 + 4 "· x 2 - x 3 + 4 · x 2" x 4 \u003d 3 x 2 · x 2 - (x 3 - 4) · 2 xx 4 \u003d x 3 - 8 x 3

We learned that the derived function would exist at all points of segments [1; 4] and [- 4; - one ] .

Now we need to define stationary points of function. We will do it with the equation x 3 - 8 x 3 \u003d 0. He has only one valid root equal to 2. It will be a stationary point of function and will fall into the first segment [1; four ] .

Calculate the values \u200b\u200bof the function at the ends of the first segment and at this point, i.e. For x \u003d 1, x \u003d 2 and x \u003d 4:

y (1) \u003d 1 3 + 4 1 2 \u003d 5 y (2) \u003d 2 3 + 4 2 2 \u003d 3 y (4) \u003d 4 3 + 4 4 2 \u003d 4 1 4

We obtained that the greatest value of the function M a x y x ∈ [1; 4] \u003d y (2) \u003d 3 will be achieved at x \u003d 1, and the smallest M i n y x ∈ [1; 4] \u003d y (2) \u003d 3 - at x \u003d 2.

The second segment does not include a single stationary point, so we need to calculate the values \u200b\u200bof the function only at the ends of the specified segment:

y (- 1) \u003d (- 1) 3 + 4 (- 1) 2 \u003d 3

It means M a x y x ∈ [- 4; - 1] \u003d y (- 1) \u003d 3, m i n y x ∈ [- 4; - 1] \u003d y (- 4) \u003d - 3 3 4.

Answer:For a segment [1; 4] - M a x y x ∈ [1; 4] \u003d y (2) \u003d 3, m i n y x ∈ [1; 4] \u003d y (2) \u003d 3, for the segment [- 4; - 1] - m a x y x ∈ [- 4; - 1] \u003d y (- 1) \u003d 3, m i n y x ∈ [- 4; - 1] \u003d y (- 4) \u003d - 3 3 4.

See Figure:

Before studying this method, we advise you to repeat how to calculate the one-sided limit and the limit on infinity, as well as learn the main methods of their stay. To find the most and / or the smallest value of the function on an outdoor or infinite interval, perform the following steps.

1. First you need to check whether a given interval will be a subset of the area of \u200b\u200bdefinition of this function.
2. We define all the points that are contained in the desired interval and in which there is no first derivative. Usually they have functions where the argument is concluded in the module sign, and in power functions with a fractional rational indicator. If these points are absent, then you can move to the next step.
3. Now we define what stationary points will fall at a given gap. First, equalize the derivative to 0, solve the equation and select the right roots. If we do not have a single stationary point or they do not fall into a given interval, then immediately go to further actions. They are determined by the view of the interval.

• If the interval has a form [a; b), then we need to calculate the value of the function at the point x \u003d A and the one-sided limit of Lim X → B - 0 F (x).
• If the interval has a form (a; b], then we need to calculate the value of the function at the point x \u003d b and the one-sided limit of Lim X → A + 0 F (X).
• If the interval has a form (a; b), then we need to calculate the one-sided limits of Lim X → B - 0 F (x), Lim X → A + 0 F (x).
• If the interval has a form [a; + ∞), then it is necessary to calculate the value at the point x \u003d a and the limit on the plus of infinity Lim X → + ∞ F (x).
• If the interval looks like (- ∞; b], we calculate the value at the point x \u003d B and the limit for minus the infinity of Lim X → - ∞ F (x).
• If - ∞; b, then we consider the one-sided limit Lim X → B - 0 F (x) and the limit for minus infinity Lim X → - ∞ F (x)
• If - ∞; + ∞, we consider the limits to minus and plus infinity Lim X → + ∞ F (x), Lim X → - ∞ F (x).

1. At the end, it is necessary to conclude based on the functions obtained and limits. There are many options here. So, if one-sided limit is minus infinity or plus infinity, it is immediately clear that nothing can be said about the smallest and greatest value of the function. Below we will analyze one typical example. Detailed descriptions will help you understand what. If necessary, you can return to Figures 4 - 8 in the first part of the material.

Condition: The function y \u003d 3 E 1 x 2 + x - 6 - 4 is given. Calculate its largest and smallest value in the intervals - ∞; - 4, - ∞; - 3, (- 3; 1], (- 3; 2), [1; 2), 2; + ∞, [4; + ∞).

Decision

First of all, we find the field definition area. In the denoter, the fraci is a square three-melan, which should not contact 0:

x 2 + x - 6 \u003d 0 d \u003d 1 2 - 4 · 1 · (- 6) \u003d 25 x 1 \u003d - 1 - 5 2 \u003d - 3 x 2 \u003d - 1 + 5 2 \u003d 2 ⇒ D (y): x ∈ (- ∞; - 3) ∪ (- 3; 2) ∪ (2; + ∞)

We obtained the field of defining the function to which all the intervals specified in the condition belong.

Now fulfill the differentiation of the function and get:

y "\u003d 3 E 1 x 2 + x - 6 - 4" \u003d 3 · E 1 x 2 + x - 6 "\u003d 3 · E 1 x 2 + x - 6 · 1 x 2 + x - 6" \u003d \u003d 3 · E 1 x 2 + x - 6 · 1 "· x 2 + x - 6 - 1 · x 2 + x - 6" (x 2 + x - 6) 2 \u003d - 3 · (2 \u200b\u200bx + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2

Consequently, derivatives exist throughout its definition.

Let us turn to the finding of stationary points. The derivative refers to 0 at x \u003d - 1 2. This is a stationary point that is in the intervals (- 3; 1] and (- 3; 2).

Calculate the value of the function at x \u003d - 4 for the gap (- ∞; - 4], as well as the limit for minus infinity:

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 \u003d 3 E 0 - 4 \u003d - 1

Since 3 E 1 6 - 4\u003e - 1, therefore, Maxyx ∈ (- ∞; - 4] \u003d y (- 4) \u003d 3 E 1 6 - 4. It does not give us the opportunity to uniquely determine the smallest value of the function. We can only It is a conclusion that below is a limit - 1, since it is precisely to this value that the function is approaching asymptotically for minus infinity.

A feature of the second interval is that there is not a single stationary point and a single strict border. Consequently, we will not be able to calculate the greatest nor the smallest function value. Having determined the limit for minus infinity and when the argument is designed to - 3 on the left side, we only get the interval of values:

lim X → - 3 - 0 3 E 1 x 2 + x - 6 - 4 \u003d Lim X → - 3 - 0 3 E 1 (x + 3) (x - 3) - 4 \u003d 3 E 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 \u003d \u003d 3 E 1 (+ 0) - 4 \u003d 3 E + ∞ - 4 \u003d + ∞ Lim X → - ∞ 3 E 1 x 2 + x - 6 - 4 \u003d 3 E 0 - 4 \u003d - 1

It means that the values \u200b\u200bof the function will be located in the interval - 1; + ∞.

To find the most function in the third gap, we define its value in the stationary point x \u003d - 1 2, if x \u003d 1. We will also need to know the one-sided limit for the case when the argument strives for - 3 on the right side:

y - 1 2 \u003d 3 E 1 - 1 2 2 + - 1 2 - 6 - 4 \u003d 3 E 4 25 - 4 ≈ - 1. 444 y (1) \u003d 3 E 1 1 2 + 1 - 6 - 4 ≈ - 1. 644 Lim X → - 3 + 0 3 E 1 x 2 + x - 6 - 4 \u003d Lim X → - 3 + 0 3 E 1 (x + 3) (x - 2) - 4 \u003d 3 E 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 \u003d \u003d 3 E 1 (- 0) - 4 \u003d 3 E - ∞ - 4 \u003d 3 · 0 - 4 \u003d - 4

We have it turned out that the most value will adopt in the stationary point maxyx ∈ (3; 1] \u003d y - 1 2 \u003d 3 E - 4 25 - 4. As for the smallest value, it cannot be determined. All we know - This is the presence of a restriction from below to - 4.

For the interval (- 3; 2), we will take the results of the previous calculation and once again we calculate what is equal to the one-sided limit when pursuing 2 on the left side:

y - 1 2 \u003d 3 E 1 - 1 2 2 + - 1 2 - 6 - 4 \u003d 3 E - 4 25 - 4 ≈ - 1. 444 Lim X → - 3 + 0 3 E 1 x 2 + x - 6 - 4 \u003d - 4 Lim X → 2 - 0 3 E 1 x 2 + x - 6 - 4 \u003d Lim X → - 3 + 0 3 E 1 (x + 3) (x - 2) - 4 \u003d 3 E 1 (2 - 0 + 3) (2 - 0 - 2) - 4 \u003d 3 E 1 - 0 - 4 \u003d 3 E - ∞ - 4 \u003d 3 · 0 - 4 \u003d - 4

So, M a x y x ∈ (- 3; 2) \u003d y - 1 2 \u003d 3 E - 4 25 - 4, and the smallest value is not possible, and the values \u200b\u200bof the function are limited to the bottom - 4.

Based on what we did in two previous calculations, we can argue that on the interval [1; 2) The function will take the greatest value at x \u003d 1, and it is impossible to find the smallest.

On the interval (2; + ∞) the function will not reach the greatest nor the smallest value, i.e. It will take values \u200b\u200bfrom the gap - 1; + ∞.

lim X → 2 + 0 3 E 1 x 2 + x - 6 - 4 \u003d Lim X → - 3 + 0 3 E 1 (X + 3) (X - 2) - 4 \u003d 3 E 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 \u003d 3 E 1 (+ 0) - 4 \u003d 3 E + ∞ - 4 \u003d + ∞ Lim x → + ∞ 3 E 1 x 2 + x - 6 - 4 \u003d 3 E 0 - 4 \u003d - 1

Calculating what will be the value of the function at x \u003d 4, we find out that M a x y x ∈ [4; + ∞) \u003d y (4) \u003d 3 E 1 14 - 4, and the specified function on the plus of infinity will be asymptotically approaching to direct y \u003d - 1.

It is comparable to what we have turned out in each calculation, with a graph of a given function. In the figure asymptotes are shown by dotted line.

That's all we wanted to tell about finding the greatest and smallest values \u200b\u200bof the function. The sequences of the actions that we have led will help make the necessary calculations as quickly and simply. But remember that it is often useful to first find out at what periods the function will decrease, and at what increase, after which you can do further conclusions. So you can more accurately determine the greatest and smallest value of the function and justify the results obtained.

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