The largest and smallest value of the function

The largest value of a function is called the largest, the smallest value is the smallest of all its values.

A function may have only one largest and only one smallest value, or may not have any at all. Finding the largest and smallest values ​​of continuous functions is based on the following properties of these functions:

1) If in some interval (finite or infinite) the function y=f(x) is continuous and has only one extremum, and if this is the maximum (minimum), then it will be the largest (smallest) value of the function in this interval.

2) If the function f(x) is continuous on some segment , then it necessarily has the largest and smallest values ​​on this segment. These values ​​are reached either at the extremum points lying inside the segment, or at the boundaries of this segment.

To find the largest and smallest values ​​on the segment, it is recommended to use the following scheme:

1. Find the derivative.

2. Find the critical points of the function where =0 or does not exist.

3. Find the values ​​of the function at critical points and at the ends of the segment and choose from them the largest f max and the smallest f min.

When solving applied problems, in particular optimization problems, the problems of finding the largest and smallest values ​​(global maximum and global minimum) of a function on the interval X are important. To solve such problems, one should, based on the condition, choose an independent variable and express the value under study through this variable. Then find the desired maximum or minimum value of the resulting function. In this case, the interval of change of the independent variable, which can be finite or infinite, is also determined from the condition of the problem.

Example. The tank, which has the shape of a rectangular parallelepiped with a square bottom, open at the top, must be tinned inside with tin. What should be the dimensions of the tank with a capacity of 108 liters. water so that the cost of its tinning is the least?

Solution. The cost of coating the tank with tin will be the lowest if, for a given capacity, its surface is minimal. Denote by a dm - the side of the base, b dm - the height of the tank. Then the area S of its surface is equal to

AND

The resulting relation establishes the relationship between the surface area of ​​the tank S (function) and the side of the base a (argument). We investigate the function S for an extremum. Find the first derivative, equate it to zero and solve the resulting equation:

Hence a = 6. (a) > 0 for a > 6, (a)< 0 при а < 6. Следовательно, при а = 6 функция S имеет минимум. Если а = 6, то b = 3. Таким образом, затраты на лужение резервуара емкостью 108 литров будут наименьшими, если он имеет размеры 6дм х 6дм х 3дм.

Example. Find the largest and smallest values ​​of a function in between.

Solution: The specified function is continuous on the entire number axis. Function derivative

Derivative at and at . Let's calculate the values ​​of the function at these points:

.

The function values ​​at the ends of the given interval are equal to . Therefore, the largest value of the function is at , the smallest value of the function is at .

Questions for self-examination

1. Formulate L'Hopital's rule for disclosure of uncertainties of the form . List the different types of uncertainties for which L'Hospital's rule can be used.

2. Formulate signs of increasing and decreasing function.

3. Define the maximum and minimum of a function.

4. Formulate the necessary condition for the existence of an extremum.

5. What values ​​of the argument (what points) are called critical? How to find these points?

6. What are sufficient signs of the existence of an extremum of a function? Outline a scheme for studying a function for an extremum using the first derivative.

7. Outline the scheme for studying the function for an extremum using the second derivative.

8. Define convexity, concavity of a curve.

9. What is the inflection point of a function graph? Specify how to find these points.

10. Formulate the necessary and sufficient signs of convexity and concavity of the curve on a given segment.

11. Define the asymptote of the curve. How to find the vertical, horizontal and oblique asymptotes of a function graph?

12. Outline the general scheme for researching a function and constructing its graph.

13. Formulate a rule for finding the largest and smallest values ​​of a function on a given segment.

In this article I will talk about how to apply the ability to find to the study of a function: to find its largest or smallest value. And then we will solve several problems from Task B15 from the Open Task Bank for .

As usual, let's start with the theory first.

At the beginning of any study of a function, we find it

To find the largest or smallest value of the function, you need to investigate on which intervals the function increases and on which it decreases.

To do this, you need to find the derivative of the function and study its intervals of sign constancy, that is, the intervals on which the derivative retains its sign.

The intervals on which the derivative of a function is positive are intervals of increasing function.

The intervals on which the derivative of a function is negative are intervals of decreasing function.

one . Let's solve task B15 (No. 245184)

To solve it, we will follow the following algorithm:

a) Find the domain of the function

b) Find the derivative of the function .

c) Set it equal to zero.

d) Let us find the intervals of constant sign of the function.

e) Find the point at which the function takes the greatest value.

f) Find the value of the function at this point.

I tell the detailed solution of this task in the VIDEO LESSON:

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2. Let's solve task B15 (No. 282862)

Find the largest value of a function on the segment

It is obvious that the function takes the greatest value on the segment at the maximum point, at x=2. Find the value of the function at this point:

Answer: 5

3 . Let's solve task B15 (No. 245180):

Find the largest value of a function

1.title="(!LANG:ln5>0">, , т.к. title="5>1">, поэтому это число не влияет на знак неравенства.!}

2. Since the scope of the original function title="(!LANG:4-2x-x^2>0">, следовательно знаменатель дроби всегда больще нуля и дробь меняет знак только в нуле числителя.!}

3. The numerator is zero at . Let's check if the ODZ belongs to the function. To do this, check if the condition title="(!LANG:4-2x-x^2>0"> при .!}

Title="4-2(-1)-((-1))^2>0">,

so the point belongs to the ODZ of the function

We examine the sign of the derivative to the right and left of the point:

We see that the function takes the greatest value at the point . Now let's find the value of the function at :

Note 1. Note that in this problem we did not find the domain of the function: we only fixed the constraints and checked whether the point at which the derivative is equal to zero belongs to the domain of the function. In this problem, this turned out to be enough. However, this is not always the case. It depends on the task.

Remark 2. When studying the behavior of a complex function, one can use the following rule:

• if the outer function of a compound function is increasing, then the function takes on its greatest value at the same point at which the inner function takes on its greatest value. This follows from the definition of an increasing function: the function increases on the interval I if the larger value of the argument from this interval corresponds to the larger value of the function.
• if the outer function of a complex function is decreasing, then the function takes on the largest value at the same point at which the inner function takes on the smallest value . This follows from the definition of a decreasing function: the function decreases on the interval I if the larger value of the argument from this interval corresponds to the smaller value of the function

In our example, the outer function - increases over the entire domain of definition. Under the sign of the logarithm is an expression - a square trinomial, which, with a negative senior coefficient, takes the largest value at the point . Next, we substitute this value of x into the equation of the function and find its largest value.

Let the function y=f(X) continuous on the segment [ a, b]. As is known, such a function reaches its maximum and minimum values ​​on this segment. The function can take these values ​​either at an interior point of the segment [ a, b], or on the boundary of the segment.

To find the largest and smallest values ​​of a function on the segment [ a, b] necessary:

1) find the critical points of the function in the interval ( a, b);

2) calculate the values ​​of the function at the found critical points;

3) calculate the values ​​of the function at the ends of the segment, that is, for x=a and x = b;

4) from all the calculated values ​​of the function, choose the largest and smallest.

Example. Find the largest and smallest values ​​of a function

on the segment.

Finding critical points:

These points lie inside the segment ; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;

at the point x= 3 and at the point x= 0.

Investigation of a function for convexity and an inflection point.

Function y = f (x) called convexup in between (a, b) , if its graph lies under a tangent drawn at any point of this interval, and is called convex down (concave) if its graph lies above the tangent.

The point at the transition through which the convexity is replaced by concavity or vice versa is called inflection point.

Algorithm for studying for convexity and inflection point:

1. Find the critical points of the second kind, that is, the points at which the second derivative is equal to zero or does not exist.

2. Put critical points on the number line, breaking it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upwards, if, then the function is convex downwards.

3. If, when passing through a critical point of the second kind, it changes sign and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find its ordinate.

Asymptotes of the graph of a function. Investigation of a function into asymptotes.

Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point of the graph to this line tends to zero with an unlimited removal of the graph point from the origin.

There are three types of asymptotes: vertical, horizontal and inclined.

Definition. Direct called vertical asymptote function graph y = f(x), if at least one of the one-sided limits of the function at this point is equal to infinity, that is

where is the discontinuity point of the function, that is, it does not belong to the domain of definition.

Example.

D( y) = (‒ ∞; 2) (2; + ∞)

x= 2 - breaking point.

Definition. Straight y=A called horizontal asymptote function graph y = f(x) at , if

Example.

x
y

Definition. Straight y=kx +b (k≠ 0) is called oblique asymptote function graph y = f(x) at , where

General scheme for the study of functions and plotting.

Function research algorithmy = f(x) :

1. Find the domain of the function D (y).

2. Find (if possible) the points of intersection of the graph with the coordinate axes (with x= 0 and at y = 0).

3. Investigate for even and odd functions ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).

4. Find the asymptotes of the graph of the function.

5. Find intervals of monotonicity of the function.

6. Find the extrema of the function.

7. Find the intervals of convexity (concavity) and inflection points of the graph of the function.

8. On the basis of the conducted research, construct a graph of the function.

Example. Investigate the function and plot its graph.

1) D (y) =

x= 4 - breaking point.

2) When x = 0,

(0; – 5) – point of intersection with oy.

At y = 0,

3) y(‒ x)= general function (neither even nor odd).

4) We investigate for asymptotes.

a) vertical

b) horizontal

c) find oblique asymptotes where

‒oblique asymptote equation

5) In this equation, it is not required to find intervals of monotonicity of the function.

6)

These critical points partition the entire domain of the function on the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the obtained results in the form of the following table.

With this service, you can find the largest and smallest value of a function one variable f(x) with the design of the solution in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables . You can also find the intervals of increase and decrease of the function.

Find the largest and smallest value of a function

y=

on the segment [ ;]

Include Theory

Function entry rules:

A necessary condition for an extremum of a function of one variable

The equation f "0 (x *) \u003d 0 is a necessary condition for the extremum of a function of one variable, i.e. at the point x * the first derivative of the function must vanish. It selects stationary points x c at which the function does not increase and does not decrease .

A sufficient condition for an extremum of a function of one variable

Let f 0 (x) be twice differentiable with respect to x belonging to the set D . If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then the point x * is the point of the local (global) minimum of the function.

If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

That point x * is a local (global) maximum.

Example #1. Find the largest and smallest values ​​of the function: on the segment .
Solution.

The critical point is one x 1 = 2 (f'(x)=0). This point belongs to the segment . (The point x=0 is not critical, since 0∉).
We calculate the values ​​of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 for x=2; f max =9 at x=1

Example #2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let us find the critical points: 1-cos(x)=2, cos(x)=1, x=± π / 3 +2πk, k∈Z. We find y''=2sin(x), calculate , so x= π / 3 +2πk, k∈Z are the minimum points of the function; , so x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example #3. Investigate the extremum function in the neighborhood of the point x=0.
Solution. Here it is necessary to find the extrema of the function. If the extremum x=0 , then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides, the derivative changes sign. At these points, one has to apply other methods to study functions to an extremum.