Option 1. at

1. Graph of a function y=f(x) shown in the figure.

Specify the largest value of this function 1

on the segment [ a; b]. a 0 1 b x

1) 2,5; 2) 3; 3) 4; 4) 2.

https://pandia.ru/text/78/524/images/image003_127.gif" width="242" height="133 src="> 1) -4; 2) -2; 3) 4; 4) 2.

4. Functions y=f(x) set on the segment [ a; b]. at

The figure shows a graph of its derivative

y=f ´(x). Explore for extremes 1 b

function y=f(x). Please indicate the quantity in your answer. a 0 1 x

minimum points.

1) 6; 2) 7; 3) 4;

5. Find the largest value of a function y \u003d -2x2 + 8x -7.

1) -2; 2) 7; 3) 1;

6. Find the smallest value of a function on the segment .

1) https://pandia.ru/text/78/524/images/image005_87.gif" width="17" height="48 src=">.

7. Find the smallest value of a function y=|2x+3| - .

1) - https://pandia.ru/text/78/524/images/image006_79.gif" width="17" height="47"> ; 4) - .

https://pandia.ru/text/78/524/images/image009_67.gif" width="144" height="33 src="> has a minimum at the point xo=1.5?

1) 5; 2) -6; 3) 4; 4) 6.at

9. Specify the largest value of the function y=f(x) ,

1 x

0 1

1) 2,5; 2) 3; 3) -3;

y=lg(100 – x2 ).

1) 10 ; 2) 100 ; 3) 2 ; 4) 1 .

11. Find the smallest value of a function y=2sin-1.

1) -1 ; 2) -3 ; 3) -2 ; 4) - .

Test 14 The largest (smallest) value of the function.

https://pandia.ru/text/78/524/images/image013_44.gif" width="130" height="115 src=">1. Graph of the function y=f(x) shown in the figure.

Specify the smallest value of this function 1

on the segment [ a; b]. a b

0 1 x

1) 0; 2) - 4 ,5; 3) -2; 4) - 3.

2. at The figure shows a graph of the function y=f(x).

How many maximum points does the function have?

1

0 1 x 1) 5; 2) 6; 3) 4; 4) 1.

3. At what point is the function y \u003d 2x2 + 24x -25 takes on the smallest value?

https://pandia.ru/text/78/524/images/image018_37.gif" width="76" height="48"> on the segment [-3;-1].

1) - https://pandia.ru/text/78/524/images/image020_37.gif" width="17" height="47 src=">; 2); 4) - 5.

https://pandia.ru/text/78/524/images/image022_35.gif" width="135" height="33 src="> has a minimum at the point xo = -2?

; 2) -6;; 4) 6.at

9. Specify the smallest value of the function y=f(x) ,

whose graph is shown in the figure. 1 x

0 1

1) -1,5; 2) -1; 3) -3;

10. Find the largest value of a function y=log11 (121 – x2 ).

1) 11;; 3) 1;

11. Find the largest value of a function y=2cos+3.

1) 5 ; 2) 3 ; 3) 2 ; 4) .

Answers :

And to solve it, you need minimal knowledge of the topic. The next academic year is ending, everyone wants to go on vacation, and in order to bring this moment closer, I immediately get down to business:

Let's start with the area. The area referred to in the condition is limited closed set of points in the plane. For example, a set of points bounded by a triangle, including the ENTIRE triangle (if from borders“Poke out” at least one point, then the area will no longer be closed). In practice, there are also areas of rectangular, round and slightly more complex shapes. It should be noted that in the theory of mathematical analysis, strict definitions are given limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and more is not needed now.

The flat area is standardly denoted by the letter , and, as a rule, is specified analytically - by several equations (not necessarily linear); less often inequalities. A typical verbal turnover: "closed arealimited by lines".

An integral part of the task under consideration is the construction of the area on the drawing. How to do it? It is necessary to draw all the listed lines (in this case 3 straight) and analyze what happened. The desired area is usually lightly hatched, and its border is highlighted with a bold line:


The same area can be set linear inequalities: , which for some reason are more often written as an enumeration list, and not system.
Since the boundary belongs to the region, then all inequalities, of course, non-strict.

And now the crux of the matter. Imagine that the axis goes straight to you from the origin of coordinates. Consider a function that continuous in each area point. The graph of this function is surface, and the small happiness is that in order to solve today's problem, we do not need to know what this surface looks like at all. It can be located above, below, cross the plane - all this is not important. And the following is important: according to Weierstrass theorems, continuous in limited closed area, the function reaches its maximum (of the "highest") and least (of the "lowest") values ​​to be found. These values ​​are achieved or in stationary points, belonging to the regionD , or at points that lie on the boundary of this region. From which follows a simple and transparent solution algorithm:

Example 1

In a limited enclosed area

Decision: First of all, you need to depict the area on the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the problem, and therefore I will immediately give the final illustration, which shows all the "suspicious" points found during the study. Usually they are put down one after another as they are found:

Based on the preamble, the decision can be conveniently divided into two points:

I) Let's find stationary points. This is a standard action that we have repeatedly performed in the lesson. about extrema of several variables:

Found stationary point belongs areas: (mark it on the drawing), which means that we should calculate the value of the function at a given point:

- as in the article The largest and smallest values ​​of a function on a segment, I will highlight the important results in bold. In a notebook, it is convenient to circle them with a pencil.

Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at the 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 if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and go to the next paragraph.

II) We investigate the border of the region.

Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subparagraphs. But it is better to do it not anyhow. From my point of view, at first it is more advantageous to consider the segments parallel to the coordinate axes, and first of all, those lying on the axes themselves. To catch the whole sequence and logic of actions, try to study the ending "in one breath":

1) Let's deal with the lower side of the triangle. To do this, we substitute directly into the function:

Alternatively, you can do it like this:

Geometrically, this means that the coordinate plane (which is also given by the equation)"cut out" from surfaces"spatial" parabola, the top of which immediately falls under suspicion. Let's find out where is she:

- the resulting value "hit" in the area, and it may well be that at the point (mark on the drawing) the function reaches the largest or smallest value in the entire area. Anyway, let's do the calculations:

Other "candidates" are, of course, the ends of the segment. Calculate the values ​​of the function at points (mark on the drawing):

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

2) To study the right side of the triangle, we substitute it into the function and “put things in order there”:

Here we immediately perform a rough check, “ringing” the already processed end of the segment:
, perfect.

The geometric situation is related to the previous point:

- the resulting value also “entered the scope of our interests”, which means that we need to calculate what the function is equal to at the appeared point:

Let's examine the second end of the segment:

Using the function , let's check:

3) Everyone probably knows how to explore the remaining side. We substitute into the function and carry out simplifications:

Line ends have already been investigated, but on the draft we still check whether we found the function correctly :
– coincided with the result of the 1st subparagraph;
– coincided with the result of the 2nd subparagraph.

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

- there is! Substituting a straight line into the equation, we get the ordinate of this “interestingness”:

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

Let's control the calculations according to the "budget" version :
, order.

And the final step: CAREFULLY look through all the "fat" numbers, I recommend even beginners to make a single list:

from which we choose the largest and smallest values. Answer write in the style of the problem of finding the largest and smallest values ​​of the function on the segment:

Just in case, I will once again comment on the geometric meaning of the result:
– here is the highest point of the surface in the region ;
- here is the lowest point of the surface in the area.

In the analyzed problem, we found 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum "exploration set" consists of three points. This happens when the function, for example, sets plane- it is quite clear that there are no stationary points, and the function can reach the maximum / minimum values ​​only at the vertices of the triangle. But there are no such examples once, twice - usually you have to deal with some kind of surface of the 2nd order.

If you solve such tasks a little, then triangles can make your head spin, and therefore I have prepared unusual examples for you to make it 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 bounded closed area.

Pay special attention to the rational order and technique of studying the border of the area, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it as you like, but in some problems, for example, in the same Example 2, there is every chance to significantly complicate your life. An approximate example of finishing assignments at the end of the lesson.

We systematize the solution algorithm, otherwise, with my diligence of a spider, it somehow got lost in a long thread of comments of the 1st example:

- At the first step, we build an area, it is desirable to shade it, and highlight the border with a thick line. During the solution, points will appear that need to be put on the drawing.

– Find stationary points and calculate the values ​​of the function only in those, which belong to the area . The obtained values ​​are highlighted in the text (for example, circled with a pencil). If the stationary point does NOT belong to the area, then we mark this fact with an icon or verbally. If there are no stationary points at all, then we draw a written conclusion that they are absent. In any case, this item cannot be skipped!

– Exploring the border area. First, it is advantageous to deal with straight lines that are parallel to the coordinate axes (if there are any). The function values ​​calculated at "suspicious" points are also highlighted. A lot has been said about the solution technique above and something else will be said below - read, re-read, delve into!

- From the selected numbers, select the largest and smallest values ​​\u200b\u200band give an answer. Sometimes it happens that the 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 that

The final examples are devoted to other useful ideas that will come in handy in practice:

Example 4

Find the largest and smallest values ​​of a function in a closed area .

I have kept the author's formulation, in which the area is given as a double inequality. This condition can be written in an equivalent system or in a more traditional form for this problem:

I remind you that with non-linear we encountered inequalities on , and if you do not understand the geometric meaning of the entry, then please do not delay and clarify the situation right now ;-)

Decision, as always, begins with the construction of the area, which is a kind of "sole":

Hmm, sometimes you have to gnaw not only the granite of science ....

I) Find stationary points:

Idiot's dream system :)

The stationary point belongs to the region, namely, lies on its boundary.

And so, it’s nothing ... fun lesson went - that’s what it means to drink the right tea =)

II) We investigate the border of the region. Without further ado, let's start with the x-axis:

1) If , then

Find where the top of the parabola is:
- Appreciate such moments - "hit" right to the point, from which everything is already clear. But don't forget to check:

Let's calculate the values ​​of the function at the ends of the segment:

2) We will deal with the lower part of the “sole” “in one sitting” - without any complexes we substitute it into the function, moreover, we will only be interested in the segment:

The control:

Now this is already bringing some revival to the monotonous ride on a knurled track. Let's find the critical points:

We decide quadratic equation do you remember this one? ... However, remember, of course, otherwise you would not have read these lines =) If in the two previous examples calculations in decimal fractions were convenient (which, by the way, is rare), then here we are waiting for the usual ordinary fractions. We find the “x” roots and, using the equation, 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:

Here are the "candidates", so the "candidates"!

For a standalone solution:

Example 5

Find the smallest and largest values ​​of a function in a closed area

An entry with curly braces reads like this: “a set of points such that”.

Sometimes in such examples they use Lagrange multiplier method, but the real need to use it is unlikely to arise. So, for example, if a function with the same area "de" is given, then after substitution into it - with a derivative of no difficulties; moreover, everything is drawn up in a “one line” (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are more complicated cases, where without the Lagrange function (where , for example, is the same circle equation) it's hard to get by - how hard it is to get by without a good rest!

All the best to pass the session and see you soon next season!

Solutions and answers:

Example 2: Decision: draw the area on the drawing:

From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest value of a function. What is it connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life, one has to solve the problem of optimizing some parameters. And this is the problem of finding the largest and smallest values ​​of the function.

It should be noted that the largest and smallest value of a function is usually sought on some interval X , which is either the entire domain of the function or part of the domain. The interval X itself can be a line segment, an open interval , an infinite interval .

In this article, we will talk about finding the largest and smallest values ​​of an explicitly given function of one variable y=f(x) .

Page navigation.

The largest and smallest value of a function - definitions, illustrations.

Let us briefly dwell on the main definitions.

The largest value of the function , which for any the inequality is true.

The smallest value of the function y=f(x) on the interval X is called such a value , which for any the inequality is true.

These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) value accepted in the interval under consideration with the abscissa.

Stationary points are the values ​​of the argument at which the derivative of the function vanishes.

Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its maximum (smallest) value on the interval X at one of the stationary points from this interval.

Also, a function can often take on the largest and smallest values ​​at points where the first derivative of this function does not exist, and the function itself is defined.

Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.

For clarity, we give a graphic illustration. Look at the pictures - and much will become clear.

On the segment

In the first figure, the function takes the largest (max y ) and smallest (min y ) values ​​at stationary points inside the segment [-6;6] .

Consider the case shown in the second figure. Change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.

In figure No. 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.

In the open range

In the fourth figure, the function takes the largest (max y ) and smallest (min y ) values ​​at stationary points within the open interval (-6;6) .

On the interval , no conclusions can be drawn about the largest value.

At infinity

In the example shown in the seventh figure, the function takes the largest value (max y ) at a stationary point with the abscissa x=1 , and the smallest value (min y ) is reached at the right boundary of the interval. At minus infinity, the values ​​of the function asymptotically approach y=3 .

On the interval, the function does not reach either the smallest or the largest value. As x=2 tends to the right, the function values ​​tend to minus infinity (the straight line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values ​​asymptotically approach y=3 . A graphic illustration of this example is shown in Figure 8.

Algorithm for finding the largest and smallest values ​​of a continuous function on the segment .

We write an algorithm that allows us to find the largest and smallest value of a function on a segment.

1. We find the domain of the function and check if it contains the entire segment .
2. We find all points at which the first derivative does not exist and which are contained in the segment (usually such points occur in functions with an argument under the module sign and in power functions with a fractional-rational exponent). If there are no such points, then go to the next point.
3. We determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them fall into the segment, then go to the next step.
4. We calculate the values ​​of the function at the selected stationary points (if any), at points where the first derivative does not exist (if any), and also at x=a and x=b .
5. From the obtained values ​​of the function, we select the largest and smallest - they will be the desired maximum and smallest values ​​of the function, respectively.

Let's analyze the algorithm when solving an example for finding the largest and smallest values ​​of a function on a segment.

Example.

Find the largest and smallest value of a function

• on the segment;
• on the interval [-4;-1] .

Decision.

The domain of the function is the entire set of real numbers, except for zero, that is, . Both segments fall within the domain of definition.

We find the derivative of the function with respect to:

Obviously, the derivative of the function exists at all points of the segments and [-4;-1] .

Stationary points are determined from the equation . The only real root is x=2 . This stationary point falls into the first segment.

For the first case, we calculate the values ​​of the function at the ends of the segment and at a stationary point, that is, for x=1 , x=2 and x=4 :

Therefore, the largest value of the function is reached at x=1 , and the smallest value – at x=2 .

For the second case, we calculate the values ​​of the function only at the ends of the segment [-4;-1] (since it does not contain a single stationary point):

Decision.

Let's start with the scope of the function. The square trinomial in the denominator of a fraction must not vanish:

It is easy to check that all intervals from the condition of the problem belong to the domain of the function.

Let's differentiate the function:

Obviously, the derivative exists on the entire domain of the function.

Let's find stationary points. The derivative vanishes at . This stationary point falls within the intervals (-3;1] and (-3;2) .

And now you can compare the results obtained at each point with the graph of the function. The blue dotted lines indicate the asymptotes.

This can end with finding the largest and smallest value of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest value of the function on any interval. This gives a clearer picture and a rigorous justification of the results.

In many problems, it is required to calculate the maximum or minimum value of a quadratic function. The maximum or minimum can be found if the original function is written in standard form: or through the coordinates of the parabola vertex: f (x) = a (x − h) 2 + k (\displaystyle f(x)=a(x-h)^(2)+k). Moreover, the maximum or minimum of any quadratic function can be calculated using mathematical operations.

Steps

The quadratic function is written in standard form

Write the function in standard form. A quadratic function is a function whose equation includes a variable x 2 (\displaystyle x^(2)). The equation may or may not include a variable x (\displaystyle x). If an equation includes a variable with an exponent greater than 2, it does not describe a quadratic function. If necessary, bring like terms and rearrange them to write the function in standard form.

The graph of a quadratic function is a parabola. The branches of a parabola point up or down. If the coefficient a (\displaystyle a) with a variable x 2 (\displaystyle x^(2)) a (\displaystyle a)

Calculate -b/2a. Meaning − b 2 a (\displaystyle -(\frac (b)(2a))) is the coordinate x (\displaystyle x) top of the parabola. If the quadratic function is written in the standard form a x 2 + b x + c (\displaystyle ax^(2)+bx+c), use the coefficients for x (\displaystyle x) and x 2 (\displaystyle x^(2)) in the following way:

• In function coefficients a = 1 (\displaystyle a=1) and b = 10 (\displaystyle b=10)
• As a second example, consider the function . Here a = − 3 (\displaystyle a=-3) and b = 6 (\displaystyle b=6). Therefore, calculate the x-coordinate of the top of the parabola as follows:

1. Find the corresponding value of f(x). Substitute the found value of "x" into the original function to find the corresponding value of f(x). This is how you find the minimum or maximum of the function.

   • In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) you calculated that the x-coordinate of the top of the parabola is x = − 5 (\displaystyle x=-5). In the original function, instead of x (\displaystyle x) substitute − 5 (\displaystyle -5)
   • In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) you found that the x-coordinate of the vertex of the parabola is x = 1 (\displaystyle x=1). In the original function, instead of x (\displaystyle x) substitute 1 (\displaystyle 1) to find its maximum value:

2. Write down the answer. Reread the condition of the problem. If you need to find the coordinates of the vertex of the parabola, write down both values ​​in your answer x (\displaystyle x) and y (\displaystyle y)(or f (x) (\displaystyle f(x))). If you need to calculate the maximum or minimum of a function, write down only the value in your answer y (\displaystyle y)(or f (x) (\displaystyle f(x))). Look again at the sign of the coefficient a (\displaystyle a) to check if you calculated the maximum or minimum.

   The quadratic function is written in terms of the coordinates of the vertex of the parabola

   1. Write the quadratic function in terms of the coordinates of the vertex of the parabola. Such an equation has the following form:

      Determine the direction of the parabola. To do this, look at the sign of the coefficient a (\displaystyle a). If the coefficient a (\displaystyle a) positive, the parabola is directed upwards. If the coefficient a (\displaystyle a) negative, the parabola is pointing down. For example:

      Find the minimum or maximum value of the function. If the function is written in terms of the coordinates of the parabola vertex, the minimum or maximum is equal to the value of the coefficient k (\displaystyle k). In the examples above:

      Find the coordinates of the vertex of the parabola. If in the problem it is required to find the vertex of the parabola, its coordinates are (h , k) (\displaystyle (h,k)). Note that when a quadratic function is written in terms of the coordinates of the parabola vertex, the subtraction operation must be enclosed in brackets (x − h) (\displaystyle (x-h)), so the value h (\displaystyle h) taken with the opposite sign.

   How to calculate the minimum or maximum using mathematical operations

   Let us first consider the standard form of the equation. Write the quadratic function in standard form: f (x) = a x 2 + b x + c (\displaystyle f(x)=ax^(2)+bx+c). If necessary, bring like terms and rearrange them to get the standard equation.

   Find the first derivative. The first derivative of a quadratic function, which is written in standard form, is equal to f ′ (x) = 2 a x + b (\displaystyle f^(\prime )(x)=2ax+b).

   Set the derivative to zero. Recall that the derivative of a function is equal to the slope of the function at a certain point. At the minimum or maximum, the slope is zero. Therefore, to find the minimum or maximum value of a function, the derivative must be equated to zero. In our example:

Sometimes in problems B15 there are "bad" functions for which it is difficult to find the derivative. Previously, this was only on probes, but now these tasks are so common that they can no longer be ignored when preparing for this exam.

In this case, other tricks work, one of which is - monotone.

The function f (x) is called monotonically increasing on the segment, if for any points x 1 and x 2 of this segment the following is true:

x 1< x 2 ⇒ f (x 1) < f (x2).

The function f (x) is called monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following is true:

x 1< x 2 ⇒ f (x 1) > f( x2).

In other words, for an increasing function, the larger x is, the larger f(x) is. For a decreasing function, the opposite is true: the more x , the smaller f(x).

For example, the logarithm increases monotonically if the base a > 1 and decreases monotonically if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.

f (x) = log a x (a > 0; a ≠ 1; x > 0)

The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:

The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, показательная функция определена для всех чисел, а не только для x > 0:

f (x) = a x (a > 0)

Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where monotony is broken.

All these functions are never found in their pure form. Polynomials, fractions and other nonsense are added to them, because of which it becomes difficult to calculate the derivative. What happens in this case - now we will analyze.

Parabola vertex coordinates

Most often, the function argument is replaced with square trinomial of the form y = ax 2 + bx + c . Its graph is a standard parabola, in which we are interested in:

1. Parabola branches - can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
2. The vertex of a parabola is the extremum point of a quadratic function, at which this function takes its smallest (for a > 0) or largest (a< 0) значение.

Of greatest interest is top of a parabola, the abscissa of which is calculated by the formula:

So, we have found the extremum point of the quadratic function. But if the original function is monotonic, for it the point x 0 will also be an extremum point. Thus, we formulate the key rule:

The extremum points of the square trinomial and the complex function it enters into coincide. Therefore, you can look for x 0 for a square trinomial, and forget about the function.

From the above reasoning, it remains unclear what kind of point we get: a maximum or a minimum. However, the tasks are specifically designed so that it does not matter. Judge for yourself:

1. There is no segment in the condition of the problem. Therefore, it is not required to calculate f(a) and f(b). It remains to consider only the extremum points;
2. But there is only one such point - this is the top of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.

Thus, the solution of the problem is greatly simplified and reduced to just two steps:

1. Write out the parabola equation y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a;
2. Find the value of the original function at this point: f (x 0). If there are no additional conditions, this will be the answer.

At first glance, this algorithm and its justification may seem complicated. I deliberately do not post a "bare" solution scheme, since the thoughtless application of such rules is fraught with errors.

Consider the real tasks from the trial exam in mathematics - this is where this technique is most common. At the same time, we will make sure that in this way many problems of B15 become almost verbal.

Under the root is a quadratic function y \u003d x 2 + 6x + 13. The graph of this function is a parabola with branches up, since the coefficient a \u003d 1\u003e 0.

Top of the parabola:

x 0 \u003d -b / (2a) \u003d -6 / (2 1) \u003d -6 / 2 \u003d -3

Since the branches of the parabola are directed upwards, at the point x 0 \u003d −3, the function y \u003d x 2 + 6x + 13 takes on the smallest value.

The root is monotonically increasing, so x 0 is the minimum point of the entire function. We have:

Task. Find the smallest value of the function:

y = log 2 (x 2 + 2x + 9)

Under the logarithm is again a quadratic function: y \u003d x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.

Top of the parabola:

x 0 \u003d -b / (2a) \u003d -2 / (2 1) \u003d -2/2 \u003d -1

So, at the point x 0 = −1, the quadratic function takes on the smallest value. But the function y = log 2 x is monotone, so:

y min = y (−1) = log 2 ((−1) 2 + 2 (−1) + 9) = ... = log 2 8 = 3

The exponent is a quadratic function y = 1 − 4x − x 2 . Let's rewrite it in normal form: y = −x 2 − 4x + 1.

Obviously, the graph of this function is a parabola, branches down (a = −1< 0). Поэтому вершина будет точкой максимума:

x 0 = −b /(2a ) = −(−4)/(2 (−1)) = 4/(−2) = −2

The original function is exponential, it is monotone, so the largest value will be at the found point x 0 = −2:

An attentive reader will surely notice that we did not write out the area of ​​\u200b\u200bpermissible values ​​of the root and logarithm. But this was not required: inside there are functions whose values ​​are always positive.

Consequences from the scope of a function

Sometimes, to solve problem B15, it is not enough just to find the vertex of the parabola. The desired value may lie at the end of the segment, but not at the extremum point. If the task does not specify a segment at all, look at tolerance range original function. Namely:

Pay attention again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how it works with specific examples:

Task. Find the largest value of the function:

Under the root is again a quadratic function: y \u003d 3 - 2x - x 2. Its graph is a parabola, but branches down since a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический квадратный корень из отрицательного числа не существует.

We write out the area of ​​​​permissible values ​​​​(ODZ):

3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; one]

Now find the vertex of the parabola:

x 0 = −b /(2a ) = −(−2)/(2 (−1)) = 2/(−2) = −1

The point x 0 = −1 belongs to the ODZ segment - and this is good. Now we consider the value of the function at the point x 0, as well as at the ends of the ODZ:

y(−3) = y(1) = 0

So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.

Task. Find the smallest value of the function:

y = log 0.5 (6x - x 2 - 5)

Inside the logarithm there is a quadratic function y \u003d 6x - x 2 - 5. This is a parabola with branches down, but there cannot be negative numbers in the logarithm, so we write out the ODZ:

6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)

Please note: the inequality is strict, so the ends do not belong to the ODZ. In this way, the logarithm differs from the root, where the ends of the segment suit us quite well.

Looking for the vertex of the parabola:

x 0 = −b /(2a ) = −6/(2 (−1)) = −6/(−2) = 3

The top of the parabola fits along the ODZ: x 0 = 3 ∈ (1; 5). But since the ends of the segment do not interest us, we consider the value of the function only at the point x 0:

y min = y (3) = log 0.5 (6 3 − 3 2 − 5) = log 0.5 (18 − 9 − 5) = log 0.5 4 = −2