In task B14 from the USE in mathematics, you need to find the smallest or largest value of a function of one variable. This is a rather trivial task from mathematical analysis, and it is for this reason that every graduate can and should learn how to solve it normally. high school. Let's analyze a few examples that schoolchildren solved at the diagnostic work in mathematics, which took place in Moscow on December 7, 2011.

Depending on the interval on which you want to find the maximum or minimum value of the function, one of the following standard algorithms is used to solve this problem.

I. Algorithm for finding the largest or smallest value of a function on a segment:

• Find the derivative of a function.
• Select from the points suspected of an extremum those that belong to a given segment and the domain of the function.
• Calculate values functions(not a derivative!) at these points.
• Among the obtained values, choose the largest or smallest, it will be the desired one.

Example 1 Find the smallest value of a function
y = x 3 – 18x 2 + 81x+ 23 on the segment .

Solution: we act according to the algorithm for finding the smallest value of a function on a segment:

• The scope of the function is not limited: D(y) = R.
• The derivative of the function is: y' = 3x 2 – 36x+ 81. The scope of the derivative of a function is also not limited: D(y') = R.
• Zeros of the derivative: y' = 3x 2 – 36x+ 81 = 0, so x 2 – 12x+ 27 = 0, whence x= 3 and x= 9, our interval includes only x= 9 (one point suspicious for an extremum).
• We find the value of the function at a point suspicious of an extremum and at the edges of the interval. For the convenience of calculations, we represent the function in the form: y = x 3 – 18x 2 + 81x + 23 = x(x-9) 2 +23:
  • y(8) \u003d 8 (8-9) 2 +23 \u003d 31;
  • y(9) = 9 (9-9) 2 +23 = 23;
  • y(13) = 13 (13-9) 2 +23 = 231.

So, from the obtained values, the smallest is 23. Answer: 23.

II. The algorithm for finding the largest or smallest value of a function:

• Find the scope of the function.
• Find the derivative of a function.
• Determine the points that are suspicious of an extremum (those points at which the derivative of the function vanishes, and the points at which there is no two-sided finite derivative).
• Mark these points and the domain of the function on the number line and determine the signs derivative(not functions!) on the resulting intervals.
• Define values functions(not a derivative!) at the minimum points (those points at which the sign of the derivative changes from minus to plus), the smallest of these values ​​will be the smallest value of the function. If there are no minimum points, then the function does not have a minimum value.
• Define values functions(not a derivative!) at the maximum points (those points at which the sign of the derivative changes from plus to minus), the largest of these values ​​will be the largest value of the function. If there are no maximum points, then the function does not have a maximum value.

Example 2 Find the largest value of the function.

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]. but 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]. but 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 :

petite and pretty simple task from the category of those that serve as a lifeline for a floating student. In nature, the sleepy realm of mid-July, so it's time to settle down with a laptop on the beach. Early in the morning, a sunbeam of theory began to play in order to soon focus on practice, which, despite its declared lightness, contains glass fragments in the sand. In this regard, I recommend conscientiously consider a few examples of this page. For solutions practical tasks need to be able find derivatives and understand the material of the article Intervals of monotonicity and extrema of a function.

First, briefly about the main thing. In a lesson about function continuity I gave the definition of continuity at a point and continuity on an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on a segment if:

1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.

The second paragraph deals with the so-called unilateral continuity functions at a point. There are several approaches to its definition, but I will stick to the line started earlier:

The function is continuous at a point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: . It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at that point:

Imagine that the green dots are the nails on which the magic rubber band is attached:

Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited- a hedge above, a hedge below, and our product grazes in a paddock. In this way, a function continuous on a segment is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and rigorously proved Weierstrass' first theorem.… Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled the graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Indeed, how do you know what awaits us beyond the horizon? After all, once the Earth was considered flat, so today even ordinary teleportation requires proof =)

According to second Weierstrass theorem, continuous on the segmentfunction reaches its exact top edge and his exact bottom edge .

The number is also called the maximum value of the function on the segment and denoted by , and the number - the minimum value of the function on the segment with notice .

In our case:

Note : in theory, records are common .

Roughly speaking, the greatest value is located where the most high point graphics, and the smallest - where is the lowest point.

Important! As already pointed out in the article on extrema of the function, the largest value of the function And smallest function value – NOT THE SAME, what function maximum And function minimum. So, in this example, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even the flood, in the context of the problem under consideration, this does not interest us at all. The task involves only finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore, no need to draw!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the function values ​​in critical points, that belong to this segment.

Catch one more goodie: there is no need to check a sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum not yet guaranteed what is the minimum or maximum value. The demonstration function reaches its maximum and, by the will of fate, the same number is the largest value of the function on the interval . But, of course, such a coincidence does not always take place.

So, at the first step, it is faster and easier to calculate the values ​​of the function at critical points belonging to the segment, without bothering whether they have extrema or not.

2) We calculate the values ​​of the function at the ends of the segment.

3) Among the values ​​of the function found in the 1st and 2nd paragraphs, we select the smallest and most big number, write down the answer.

We sit on the shore of the blue sea and hit the heels in shallow water:

Example 1

Find the largest and smallest value functions on the segment

Solution:
1) Calculate the values ​​of the function at critical points belonging to this segment:

Let us calculate the value of the function at the second critical point:

2) Calculate the values ​​of the function at the ends of the segment:

3) "Bold" results were obtained with exponentials and logarithms, which significantly complicates their comparison. For this reason, we will arm ourselves with a calculator or Excel and calculate the approximate values, not forgetting that:

Now everything is clear.

Answer:

Fractional-rational instance for independent solution:

Example 6

Find the maximum and minimum values ​​of a function on a segment

\(\blacktriangleright\) In order to find the largest/smallest value of a function on the segment \(\) , it is necessary to schematically depict the graph of the function on this segment.
In the problems from this subtopic, this can be done using the derivative: find the intervals of increase (\(f">0\) ) and decrease (\(f"<0\) ) функции, критические точки (где \(f"=0\) или \(f"\) не существует).

\(\blacktriangleright\) Do not forget that the function can take the maximum/smallest value not only at the internal points of the segment \(\) , but also at its ends.

\(\blacktriangleright\) The largest/smallest value of the function is the value of the coordinate \(y=f(x)\) .

\(\blacktriangleright\) The derivative of a complex function \(f(t(x))\) is searched according to the rule: \[(\Large(f"(x)=f"(t)\cdot t"(x)))\]
\[\begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x)\\ \hline \textbf(1) & c & 0\\&&\\ \textbf(2) & x^a & a\cdot x^(a-1)\\&&\\ \textbf(3) & \ln x & \dfrac1x\\&&\\ \ textbf(4) & \log_ax & \dfrac1(x\cdot \ln a)\\&&\\ \textbf(5) & e^x & e^x\\&&\\ \textbf(6) & a^x & a^x\cdot \ln a\\&&\\ \textbf(7) & \sin x & \cos x\\&&\\ \textbf(8) & \cos x & -\sin x\\ \hline \end(array) \quad \quad \quad \quad \begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x) \\ \hline \textbf(9) & \mathrm(tg)\, x & \dfrac1(\cos^2 x)\\&&\\ \textbf(10) & \mathrm(ctg)\, x & -\ ,\dfrac1(\sin^2 x)\\&&\\ \textbf(11) & \arcsin x & \dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(12) & \ arccos x & -\,\dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(13) & \mathrm(arctg)\, x & \dfrac1(1+x^2)\\ &&\\ \textbf(14) & \mathrm(arcctg)\, x & -\,\dfrac1(1+x^2)\\ \hline \end(array)\]

Task 1 #2357

Task level: Equal to the Unified State Examination

Find the smallest value of the function \(y = e^(x^2 - 4)\) on the interval \([-10; -2]\) .

ODZ: \(x\) - arbitrary.

1) \

\ So \(y" = 0\) when \(x = 0\) .

3) Let's find intervals of constant sign \(y"\) on the considered segment \([-10; -2]\) :

4) Sketch of the graph on the segment \([-10; -2]\) :

Thus, the function reaches its smallest value on \([-10; -2]\) at \(x = -2\) .

\ Total: \(1\) is the smallest value of the function \(y\) on \([-10; -2]\) .

Answer: 1

Task 2 #2355

Task level: Equal to the Unified State Examination

\(y = \sqrt(2)\cdot\sqrt(x^2 + 1)\) on the segment \([-1; 1]\) .

ODZ: \(x\) - arbitrary.

1) \

Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[\sqrt(2)\cdot\dfrac(x)(\sqrt(x^2 + 1)) = 0\qquad\Leftrightarrow\qquad x = 0\,.\] The derivative exists for any \(x\) .

2) Find the intervals of constant sign \(y"\) :

3) Let's find intervals of constant sign \(y"\) on the considered segment \([-1; 1]\) :

4) Sketch of the graph on the segment \([-1; 1]\) :

Thus, the function reaches its maximum value on \([-1; 1]\) in \(x = -1\) or in \(x = 1\) . Let's compare the values ​​of the function at these points.

\ Total: \(2\) is the largest value of the function \(y\) on \([-1; 1]\) .

Answer: 2

Task 3 #2356

Task level: Equal to the Unified State Examination

Find the smallest value of the function \(y = \cos 2x\) on the interval \(\) .

ODZ: \(x\) - arbitrary.

1) \

Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[-2\cdot \sin 2x = 0\qquad\Leftrightarrow\qquad 2x = \pi n, n\in\mathbb(Z)\qquad\Leftrightarrow\qquad x = \dfrac(\pi n)(2), n\in\mathbb(Z)\,.\] The derivative exists for any \(x\) .

2) Find the intervals of constant sign \(y"\) :

(here there is an infinite number of intervals in which the signs of the derivative alternate).

3) Let's find intervals of constancy \(y"\) on the considered segment \(\) :

4) Sketch of the graph on the segment \(\) :

Thus, the function reaches its smallest value on \(\) at \(x = \dfrac(\pi)(2)\) .

\ Total: \(-1\) is the smallest value of the function \(y\) on \(\) .

Answer: -1

Task 4 #915

Task level: Equal to the Unified State Examination

Find the largest value of a function

\(y = -\log_(17)(2x^2 - 2\sqrt(2)x + 2)\).

ODZ: \(2x^2 - 2\sqrt(2)x + 2 > 0\) . Let's decide on ODZ:

1) Denote \(2x^2-2\sqrt(2)x+2=t(x)\) , then \(y(t)=-\log_(17)t\) .

Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[-\dfrac(1)(\ln 17)\cdot\dfrac(4x-2\sqrt(2))(2x^2-2\sqrt(2)x+2) = 0\qquad\Leftrightarrow\qquad 4x-2\sqrt(2) = 0\]– on the ODZ, from where we find the root \(x = \dfrac(\sqrt(2))(2)\) . The derivative of the function \(y\) does not exist for \(2x^2-2\sqrt(2)x+2 = 0\) , but given equation negative discriminant, hence it has no solutions. In order to find the largest / smallest value of a function, you need to understand how its graph looks schematically.

2) Find the intervals of constant sign \(y"\) :

3) Graphic sketch:

Thus, the function reaches its maximum value at \(x = \dfrac(\sqrt(2))(2)\) :

\(y\left(\dfrac(\sqrt(2))(2)\right) = -\log_(17)1 = 0\),

Total: \(0\) is the largest value of the function \(y\) .

Answer: 0

Task 5 #2344

Task level: Equal to the Unified State Examination

Find the smallest value of a function

\(y = \log_(3)(x^2 + 8x + 19)\).

ODZ: \(x^2 + 8x + 19 > 0\) . Let's decide on ODZ:

1) Denote \(x^2 + 8x + 19=t(x)\) , then \(y(t)=\log_(3)t\) .

Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[\dfrac(1)(\ln 3)\cdot\dfrac(2x+8)(x^2 + 8x + 19) = 0\qquad\Leftrightarrow\qquad 2x+8 = 0\]- on the ODZ, from where we find the root \ (x \u003d -4 \) . The derivative of the function \(y\) does not exist for \(x^2 + 8x + 19 = 0\) , but this equation has a negative discriminant, therefore, it has no solutions. In order to find the largest / smallest value of a function, you need to understand how its graph looks schematically.

2) Find the intervals of constant sign \(y"\) :

3) Graphic sketch:

Thus, \(x = -4\) is the minimum point of the function \(y\) and the smallest value is reached in it:

\(y(-4) = \log_(3)3 = 1\) .

Total: \(1\) is the smallest value of the function \(y\) .

Answer: 1

Task 6 #917

Task level: More difficult than the exam

Find the largest value of a function

\(y = -e^((x^2 - 12x + 36 + 2\ln 2))\).

Often it is necessary to solve problems in which it is necessary to find the largest or smallest value from the set of those values ​​that a function takes on a segment.

Let us turn, for example, to the graph of the function f (x) \u003d 1 + 2x 2 - x 4 on the segment [-1; 2]. To work with a function, we need to plot its graph.

It can be seen from the constructed graph that the function takes the largest value on this segment, equal to 2, at the points: x = -1 and x = 1; the smallest value equal to -7, the function takes at x = 2.

The point x \u003d 0 is the minimum point of the function f (x) \u003d 1 + 2x 2 - x 4. This means that there is a neighborhood of the point x \u003d 0, for example, the interval (-1/2; 1/2) - such that in this neighborhood the function takes the smallest value at x \u003d 0. However, on a larger interval, for example, on the segment [ -one; 2], the function takes the smallest value at the end of the segment, and not at the minimum point.

Thus, in order to find the smallest value of a function on a certain segment, it is necessary to compare its values ​​at the ends of the segment and at the minimum points.

In general, suppose that the function f(x) is continuous on a segment and that the function has a derivative at every interior point of this segment.

In order to find the largest and smallest values ​​of a function on a segment, it is necessary:

1) find the values ​​of the function at the ends of the segment, i.e. numbers f(a) and f(b);

2) find the values ​​of the function at stationary points that belong to the interval (a; b);

3) choose the largest and smallest from the found values.

Let's apply the acquired knowledge in practice and consider the problem.

Find the largest and smallest values ​​of the function f (x) \u003d x 3 + x / 3 on the segment.

Solution.

1) f(1/2) = 6 1/8, f(2) = 9 ½.

2) f´(x) \u003d 3x 2 - 3 / x 2 \u003d (3x 4 - 3) / x 2, 3x 4 - 3 \u003d 0; x 1 = 1, x 2 = -1.

The interval (1/2; 2) contains one stationary point x 1 = 1, f(1) = 4.

3) Of the numbers 6 1/8, 9 ½ and 4, the largest is 9 ½, the smallest is 4.

Answer. The largest feature value is 9 ½, the smallest feature value is 4.

Often, when solving problems, it is necessary to find the largest and smallest value of a function not on a segment, but on an interval.

In practical problems, the function f(x) usually has only one stationary point on a given interval: either a maximum point or a minimum point. In these cases, the function f(x) takes the largest value in a given interval at the maximum point, and at the minimum point, the smallest value in this interval. Let's turn to the problem.

The number 36 is written as a product of two positive numbers, the sum of which is the smallest.

Solution.

1) Let the first factor be x, then the second factor is 36/x.

2) The sum of these numbers is x + 36/x.

3) According to the conditions of the problem, x is a positive number. So, the problem is reduced to finding the value of x - such that the function f (x) \u003d x + 36 / x takes the smallest value on the interval x > 0.

4) Find the derivative: f´(x) \u003d 1 - 36 / x 2 \u003d ((x + 6) (x - 6)) / x 2.

5) Stationary points x 1 = 6, x 2 = -6. On the interval x > 0, there is only one stationary point x = 6. When passing through the point x = 6, the derivative changes sign “–” to sign “+”, and therefore x = 6 is the minimum point. Consequently, the function f(x) = x + 36/x takes the smallest value on the interval x > 0 at the point x = 6 (this is the value f(6) = 12).

Answer. 36 = 6 ∙ 6.

When solving some problems where it is necessary to find the largest and smallest values ​​of a function, it is useful to use the following statement:

if the values ​​of the function f(x) are non-negative on some interval, then this function and the function (f(x)) n , where n is natural number, take the largest (smallest) value at the same point.

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