§3. The largest and smallest value of a function on a segment. Algorithm for finding the highest and lowest values ​​of a function on a segment

1. Find the domain of definition of the function and check whether it contains the entire segment.

2. Determine all stationary points falling within the segment. To do this, we find the derivative of the function, equate it to zero, solve the resulting equation and select suitable roots.

3. If there are no stationary points or none of them fall into the segment, then move on to the next point.

4. We calculate the values ​​of the function at selected stationary points (if any), as well as at x = a and x = b.

5. From the obtained function values, select the largest and smallest - they will be the ones we are looking for.

10) Sufficient condition for convexity (concavity). If the second derivative of a twice differentiable function is positive (negative) on a set X, then the function is downward (upward) convex on this set.

11) Necessary condition for inflection points. The second derivative f""(x) of a twice continuously differentiable function at the inflection point x0 is equal to zero, i.e. f""(x0) = 0.

12) Sufficient condition for inflection points. If the second derivative of a twice differentiable function changes its sign when passing through the point x0 at which f""(x0) = 0, then x0 is the inflection point of its graph.

6.Differential calculus of functions of several variables.

Partial derivatives of functions z = f(x,y) are called the limits of the ratio of increments of the function z = z(x,y) to the increment of the corresponding argument in directions Oh or OU at Δx → 0 And Δу → 0 respectively:

Partial derivative with respect to x:

When calculating, consider y = const.

Partial derivative with respect to y:

When calculating, consider x = const.

The set G of all pairs of values ​​of the arguments of a given function of two variables is called domain of definition of this function.

The function z = f(x,y) is called continuous at the point M0(x0,y0), if it is defined at this point and its neighborhood and satisfies

The number A is called limit of the function z = f(x,y) at point M0(x0,y0):

The linear (relative to delta x and delta ig) part of the total increment of the function is called full differential and is denoted by dz:

where deix and deigric are differentials of independent variables, which, by definition, are equal to the corresponding increments

Dot (x 0; y 0) called a point maximum function z = f(x; y) (x 0; y 0) For

= <δ f(x; y)≤ f(x 0; y 0).

Dot (x 0; y 0) called a point minimum function z = f(x; y) , if everywhere in the neighborhood of the point (x 0; y 0) For

= <δ f(x; y) ≥ f(x 0; y 0).

Let there be a surface given by the equation . The plane in which all tangent lines to lines on the surface passing through a given point are located , called tangent plane to the surface at point M0.

A straight line drawn through a point surfaces , perpendicular to the tangent plane is called normal to the surface.

If the surface is given by the equation , then the equation of the tangent plane to this surface at the point is written in the form: , and the equation of the normal to the surface at the same point is in the form:

Necessary conditions for differentiability: if a function f is differentiable at a point x0, then at this point it has partial derivatives with respect to all variables. if a function f is differentiable at a point x0, then it is continuous at this point.

Sufficient conditions for differentiability: Let the function f() be defined in some neighborhood of the point x0. Let a function in this neighborhood have continuous partial derivatives with respect to all variables, then the function f is differentiable at this point.

The necessary conditions existence of an extremum : or at least one partial derivative does not exist.

Sufficient conditions existence of an extremum functions of two variables:If > 0

then for a) > 0 function has a minimum ( min)

IN) < 0 the function has a maximum ( max)

If<0 That no extremum.

If= 0, then additional research using derivatives of higher orders is necessary.

Complex numbers

Definitions:

1) Complex number- expansion of the set of real numbers, usually denoted by . Any complex number can be represented as a formal sum , where and are real numbers and is the imaginary unit.

2) Writing a complex number in the form , , is called algebraic form complex number.

3) The angle (in radians) of the radius vector of the point corresponding to the number is called argument numbers and is denoted by .

4) Module of a complex number is the length of the radius vector of the corresponding point of the complex plane (or, what is the same, the distance between the point of the complex plane corresponding to this number and the origin of coordinates).

The modulus of a complex number is denoted and defined by the expression . Often denoted by the letters or . If it is a real number, then it coincides with the absolute value of this real number.

5) If a complex number, then the number is called conjugate(or complex conjugate) to (also denoted by ). On the complex plane, conjugate numbers are obtained as mirror images of each other relative to the real axis. The modulus of the conjugate number is the same as the original one, and their arguments differ in sign.

6) If the real and imaginary parts of a complex number are expressed through the modulus and argument ( , ), then any complex number except zero can be written in trigonometric forms e

7) Definition products of complex numbers is established in such a way that the numbers a + b·i and a′ + b′·i can be multiplied as algebraic binomials, and that the number i has the property i 2 =−1.

8) Let be an arbitrary natural number . nth root of a complex number z is a complex number such that .

9) Exponential form of writing complex numbers

Where is the expansion of the exponential for the case of a complex exponent.

Properties and theorems:

1) The product of two complex numbers in algebraic form is a complex number whose modulus is equal to the product of the moduli of the factors, and whose argument is equal to the sum of the arguments of the factors.

2) In order to multiply two complex numbers in trigonometric form records need to be multiplied by their modules, and the arguments added. Let , where and , where are two arbitrary complex numbers written in trigonometric form. Then .

3) Moivre's formula for complex numbers states that for any

4) In order to divide a complex number (a 1 + b 1 i) to another complex number ( a 2 + b 2 i), that is, find , you need to multiply both the numerator and the denominator by the number conjugate to the denominator.

5)

8.Integral calculus of functions of one variable.

1) Antiderivative

A function F(x) that is differentiable on a certain interval (a,b) is called antiderivative for the function f(x) on this interval if for each x (a,b) the equality is true

2) Indefinite integral

If F(x) is antiderivative for the function f(x) on a certain interval, then the expression F(x)+C is called the indefinite integral of the function f(x) and is denoted

3) Definite integral

By a definite integral of a given function f(x) on a given segment we mean the corresponding increment of its antiderivative, i.e.

4) Improper integral of a discontinuous function

Let the function f(x) be continuous a ≤x≤b and have a discontinuity point at x=b. Then the corresponding improper integral of the discontinuous function is determined by the formula

and is called convergent or divergent depending on whether the limit on the right side of the equality exists or does not exist

5) Improper integral with infinite integration interval

Let the function f(x) be continuous for a≤x≤b+∞. Then by definition

If the limit exists, then the integral on the left side of the equality is called convergent and its value is determined by the formula; otherwise the equality loses its meaning, the integral on the left is called divergent and no numerical value is assigned to it

Properties and theorems

6) Formula for integration by parts in the indefinite integral

7) Formulate the rules for integrating fractional-rational functions

1. Divide the numerator by the denominator

2. Q(x) =(x- )(x- )…

3. We expand the fraction into the sum of simple fractions; ; ; ;

The integral of fractions of types 1 and 2 is calculated by introducing the function under the sign of the differential, 3 and 4, first, a complete square is selected in the denominator.

8) Formulate the rule for integrating trigonometric functions

9) Formulate the properties of a definite integral

1. The value of the definite integral does not depend on the designation of the integration variable, i.e.

2. The definite integral with the same limits is equal to zero

3. When rearranging the limits of integration, the definite integral changes its sign to the opposite one

4. If the integration interval is divided into a finite number of partial intervals, then the definite integral taken over the interval is equal to the sum of definite integrals taken over all its partial intervals

5. The constant factor can be taken out of the sign of the definite integral

6. The definite integral of an algebraic sum of a finite number of continuous functions is equal to the same algebraic sum of definite integrals of these functions

10) Newton-Leibniz formula

If f is continuous on an interval and F is any antiderivative of it on this interval, then the equality holds

11) Formula for integration by parts in a definite integral

For brevity, we use the notation

2) Formulate the properties of the indefinite integral

1. The differential of the indefinite integral is equal to the integrand, and the derivative of the indefinite integral is equal to the integrand

2. The indefinite integral of the differential of a continuously differentiable function is equal to this function itself up to a constant term

3. A non-zero constant factor can be taken out of the sign of the indefinite integral

4. The indefinite integral of the algebraic sum of a finite number of continuous functions is equal to the same algebraic sum of the indefinite integrals of these functions

5) Change of variable in the indefinite integral

Suppose we need to find the integral. Let's introduce a new variable t, setting x= (t), where (t) is a continuous function with a continuous derivative, which has an inverse function t=Ψ(t). Then, on the right side after integration, one should make the substitution t=Ψ(x)

3) Table of integrals

Logarithms

Exponential functions

Irrational functions

Trigonometric functions

12) Change of variable in a definite integral

The function f(x) is continuous on the interval, the function x= (t) has a continuous derivative on the interval [, with a≤ (t)≤b and =a, =b

13) Calculation of the area of ​​a flat figure

Let the function f(x) be continuous on the interval. If f(x)≥0 on , then the area of ​​the curvilinear trapezoid bounded by the lines y=f(x), y=0, x=a, x=b, will be expressed using the integral:

If f(x)≤0 on , then –f(x)≥0 on . Therefore, the area S of the corresponding curvilinear trapezoid is found by the formula

In polar coordinates

The value of the function at the point max is greatest only in a certain neighborhood of this point and is not necessarily the case. the greatest value in the entire field of definition of the function. The same can be said about the minimum. In this case, they are often called local (local) max and min in contrast to absolute ones, i.e. - the largest and smallest value. throughout the region of definition. If the function f(x) is given on а,в and is continuous on it, then it reaches its maximum and minimum values ​​on it at some points. How to find them? If there are several max on a,b, then max. the value inside (if reached) matches one of them. At the same time, the function can reach its greatest value for all a,b at one of the ends.

Rule..

It is necessary to compare all min and boundary values ​​f(a) and f(b). The smallest value will be the smallest value of the function on a,b. Usually they act when they find the most. and name simpler values:

Find all critical points inside the segment a,b, calculate the values ​​of the function in them (without determining whether they have an extremum), 2) calculate the value of the function at the ends f(a) and f(b), 3) compare the obtained values ​​between is: the smallest value of these values ​​will be the smallest value of the function, the largest will be the largest on a,b.

Example:

Naiti naib. and the smallest value of the function y=na-1.2,

1. looking for critical points at (-1,2).

U"=
=0, 2x+2x 3 -2x 3 =0, 2x=0, =0. No others.

2. f(-1)=1/2, f(2)=4/5.

f(0)=0, the smallest value, f(2)=4/5.- the largest value

The following should be noted. In applied problems, the most common case is when between a and b the function y = f (x) im. only one critical point. In this case, without comparison with the boundary values, it is clear that if, incl. max, then this is the largest value of the function on а,в, if it is min, then this is the smallest value on а,в. This is important in cases where the function expression includes literal expressions and it turns out to be easier to examine the extremum than to compare the values ​​at the ends.

It is important to note that everything that has been said about finding the maximum and minimum values ​​applies to both (a, b) and the infinite interval , only in this case the values ​​​​at the ends are not taken into account.

§4. Direction of concavity of the curve and inflection point

Let the function y=f(x) im. incl. final derivative. Then she told them. at this point the tangent whose equation is y- =f "( )(X- ) or y=f( )+(x- )
.

In some neighborhood ( -The graph of the function can be located in different ways: either above the tangent, or below, or on both sides.

Definition.

They say that in t.M( ,) the curve y=f(x) is concave downward or simply concave (concave upward or convex), if for all x from some neighborhood ( - points all points of the curve are located above the tangent (below the tangent).

If in T.M the curve passes from one side of the tangent to the other, then T.M is called. inflection point of the curve.

In t. M1 - the curve is concave, M2 is convex, M3 is an inflection.

At the inflection point, the curve changes from convex to concave or vice versa. The inflection point is the boundary between the convex and concave sections of the curve.

The definition of the inflection point remains valid in the case when the tangent to the curve y = f (x) is perpendicular. axes oh, those in t. derivativef "( )=, etc. not yavl. cusp point of the curve. Unlike the cases (indicated in the drawing),

x x

where t. and x are not inflection points.

Let's find the conditions under which they. the location of a certain direction of concavity or inflection of a curve. y=f(x) in an arbitrary t.x= .

Let, for example, a curve in t.M( ,) convex. Then it is located in some neighborhood ( - of this point is below the tangent y=f( )+f "( )(X- ). Let's consider the auxiliary function(x)= f(x)-f( )-f "( )(X- ). Incl. ()=0, in-neighborhood t.
. It follows that at the point function
hasmax. So at the point ""(). But ""( )=f ""(x) and therefore incl. f ""( ).

Thus, for the curve y=f(x) to be convex at t.x0 it is necessary that f ""( ). If in t.x0 f ""( ), then incl. -max and the curve is therefore convex. Condition f ""( ) sufficient for convexity incl. .

Reasoning in a completely similar way, we obtain that the condition f ""( ) necessary for concavity at t.x0, and the condition f ""( ) sufficient for concavity.

Conclusion:

if in t. the second derivative is positivef ""( ), then the curve is curved at this point, if in t. the second derivative is negativef ""( ), then the curve is convex at this point.

The “cup” rule is convenient:

At the points of inflection there is no definite concavity or convexity, and therefore they can only be at points where f ""( )=0. But the condition f ""( ) does not yet ensure exactly that - inflection point. For example, for the curves y=x 4 and y=-x 4, incl. f ""( )=0, but in it the first curve is concave, the second is convex.

Conclusion: condition f ""( )=0 yavl. a necessary condition for the existence of an inflection, including . But, as we have seen, there can be inflections where the second derivative f""( )= silt does not exist at all.

A sufficient condition for the inflection of the curve, incl. yavl. change of sign of the second derivative f ""( ) when passing through t. . Moreover, if the 2nd derivative changes when passing through t. sign from + to -, then incl. bend with a change from concavity to convexity, Iff ""( ) changes sign from - to + when passing through t. , then incl. bend with a change from convexity to concavity..

Definition . If a curve is concave (convex) at every point of a certain interval, then it is called. concave (convex) on this interval.

The study of the function y=f(x) for convexity, concavity, and inflection points is carried out according to the following plan:

1. Find all points suspicious for inflection, for which:

a) find the second derivative, equate it to zero and find the real roots of the resulting equation,

b) find points where the finite derivative f ""(x) does not exist,

2. Examine f ""(x) for a change in sign when passing through each point suspicious of an inflection. If the sign changes, there is an inflection; if not, there is no bend.

For those points where f ""(x0)  the curve is concave, where, on the contrary, it is convex. Just as in the case of extrema, if there are a finite number of points suspicious for inflection, use the interval method.

Definition.

If a curve is convex (concave) at every point of a certain interval, it is called. convex (concave) on this interval.

Example

Examine the protrusion, concavity, i.e. inflection of the function y=x 4 -6x 2 +5. Region def. X=.

1. find y"=4x 3 -12x, y""=12x 2 -12=12(x 2 -1), y""=0, x 2 -1=0, x 1.2 =-t .suspicious for bending, no others.

The whole region def. is divided into intervals (--1), (-1,1), (1, , in each of them f ""(x) has a constant sign, because it is continuous in them. It is easy to see , that in (--1) +, in (-1,1) -, and in (1,  +. From here it is clear that in points -1 and 1 there is an inflection, and in ( -1) the graph of the function is concave, in (-1,1) it is convex, in (1,  it is concave.

LESSON PLAN No. 100

Discipline Mathematics

Speciality

Course 1 group C 153

Lesson topic: The largest and smallest values ​​of functions

Lesson type: lesson on consolidating knowledge and developing skills

Type of lesson: practical lesson

Goals:

– educational: Create an algorithm for finding the largest and smallest values ​​of a function on a segment. Carry out initial consolidation and initial control of the assimilation of the algorithm;

– developing: Develop logical thinking, computational skills;

– educational: to promote independence, self-knowledge, self-creation and self-realization in students.

Tasks:

Must know: Finding the largest and smallest values ​​of a function

Must be able to: apply acquired knowledge in practice

Formed competencies:

– general: OK 1-9

– professional: PC 1.1. – PC 4.3.

Providing classes: cards, OK

Intradisciplinary connections: lesson on the topic “The largest and smallest values ​​of a function” is associated with such topics as: “Definition of the derivative, its geometric and physical meaning”, “Derivatives of basic elementary functions”, “The second derivative, its physical meaning”, “Finding speed and acceleration using the derivative "," Differentiation of complex functions ", " Sign of constancy, increase and decrease of a function ", " Extrema of a function. Study of a function to the extremum", "Study of a function using the derivative", "Application of the derivative to the construction of graphs", "Application of the derivative to the study and construction of functions", "Convexity of the graph of a function, inflection points", "Solving exercises on the topic: "Derivative and her application"

Teaching methods: active: verbal, visual

Progress of the lesson

Organization of the lesson (3 min.).

Communicate the topic and objectives of the lesson. (4 min.)

Updating basic knowledge as a transition to mastering new knowledge. (7min.)

To study a new topic, we need to repeat the material we have covered. You will do this by completing the following tasks orally. In your notebook, write down only the answers to each item. (3 min.)

Using the graph of the function y=f(x), find:

1.Domain of definition of a function.

2. Abscissas of points at which f`(x)=0

3. Abscissas of points at which f`(x) does not exist.

4. The greatest value of the function. (Unaib.).

5. The smallest value of the function (Unaim.).

Teacher: What points are called stationary?

Student: Points at which the derivative of the function f / (x) = 0 are called stationary.

Teacher: To find stationary points you need to: find the derivative of the function f / (x) and solve the equation f / (x)= 0

Communication and assimilation of new knowledge with consolidation of acquired knowledge. (41 min.)

Algorithm for finding the smallest and largest values ​​of the continuous function y=f(x) on the segment [a; b]

find f "(x);

find points at which f "(x)=0 or f "(x) does not exist, and select from them those that lie inside the segment;

calculate the values ​​of the function y=f "(x) at the points obtained in step 2 and at the ends of the segment and select the largest and smallest from them; they will be, respectively, the largest and smallest values ​​of the function y=f(x) on the segment, which can be denoted as follows: max y(x) and min y(x).

Example.

Let's find the largest and smallest values ​​of the function on the segment.

Let's find critical points.

Since the derivative of a function is defined for any X, let's solve the equation

Consolidation of new material. Problem solving.

Option 1.

Find U max. and U name. Functions y=2-8x+6 on the segment [-1;4]

Select points belonging to the segment [-1;4]

3. Find y(-1)

Option 2.

Find U max. and U name. Functions y=+4x-3 on a segment

Find stationary points by solving the equation y´=0

Select points belonging to the segment [-3;2]

3. Find y(-3)

And at selected points in the second step

Select the largest and smallest values ​​among the values ​​found.

Solving a task from a textbook

Independent work

Option 1. Determine the largest and smallest values ​​of the function y = x 2 + 4x on the segment [-3;6].

Answer options:

a) min y(x)= -12, max y(x)= -5; b) min y(x)= -4, max y(x)= 60; c) min y(x)= -12, max y(x)= 4

[-3;6] [-3;6] [-3;6] [-3;6] [-3;6] [-3;6]

Option 2. Determine the largest and smallest values ​​of the function y = x 2 -2x on the segment.

Answer options:

a) min y(x)= -1, max y(x)= -3/4; b) min y(x)= -1, max y(x)= 8; c) min y(x)= -3/4, max y(x)= -1

Option 3. Determine the largest and smallest values ​​of the function y = 3x 2 + 6x on the segment [-2;2].

Answer options:

a) min y(x)= -4, max y(x)= 0; b) min y(x)= -20, max y(x)= 0; c) min y(x)= -3, max y(x)= 24

[-2;2] [-2;2] [-2;2] [-2;2] [-2;2] [-2;2]

Option 4. Determine the largest and smallest values ​​of the function y = 2x 2 - 2x on the segment [-1;3].

Answer options:

a) min y(x)= -0.5, max y(x)= 12; b) min y(x)= 4, max y(x)= 5; c) min y(x)= 0, max y(x)= 5

[-1;3] [-1;3] [-1;3] [-1;3] [-1;3] [-1;3]

Summing up the lesson. (5 minutes.)

What did we do in class today?

What did you like, what types of activities?

Analysis of student work, grading

Lesson reflection. (5 minutes.)

Continue the sentences:

I found out today...

I was interested in the task...

The most difficult task for me was...

I liked the lesson...

I didn't like the job...

Assignment for extracurricular independent work. (5 minutes.)

With this service you can find the largest and smallest value of a function one variable f(x) with the solution formatted 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 increasing and decreasing functions.

Rules for entering functions:

Necessary condition for the extremum of a function of one variable

Sufficient condition for the extremum of a function of one variable

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

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

If at point x * the condition is met:

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

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

Example No. 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 at x=2; f max =9 at x=1

Example No. 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's find the critical points: 1-cos(x)=2, cos(x)=½, x=± π / 3 +2πk, k∈Z. We find y’’=2sin(x), calculate , which means x= π / 3 +2πk, k∈Z are the minimum points of the function; , which means x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example No. 3. Investigate the extremum function in the vicinity 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 can 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 it is necessary to use other methods to study functions at an extremum.

From a practical point of view, the greatest interest is in using the derivative to find the largest and smallest values ​​of a function. What is this connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life we ​​have to solve problems of optimizing some parameters. And these are the tasks of finding the largest and smallest values ​​of a function.

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

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

Page navigation.

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

Let's briefly look at the main definitions.

The largest value of the function that for anyone inequality is true.

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

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

Stationary points– these are the values ​​of the argument at which the derivative of the function becomes zero.

Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. From this theorem it follows 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 largest (smallest) value on the interval X at one of the stationary points from this interval.

Also, a function can often take on its largest and smallest values ​​at points at which 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 definition of the function, or the interval X is infinite. And some functions at infinity and at the boundaries of the domain of definition can take on 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 will give a graphic illustration. Look at the pictures and a lot will become clearer.

On the segment

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

Consider the case depicted in the second figure. Let's change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest at the point with the abscissa corresponding to the right boundary of the interval.

In Figure 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.

On an open interval

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

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

At infinity

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

Over the interval, the function reaches neither the smallest nor the largest value. As x=2 approaches from the right, the function values ​​tend to minus infinity (the 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 a segment.

Let us write an algorithm that allows us to find the largest and smallest values ​​of a function on a segment.

1. We find the domain of definition of the function and check whether it contains the entire segment.
2. We find all the points at which the first derivative does not exist and which are contained in the segment (usually such points are found in functions with an argument under the modulus sign and in power functions with a fractional-rational exponent). If there are no such points, then move on to the next point.
3. We determine all stationary points falling within the segment. To do this, we equate it to zero, solve the resulting equation and select suitable roots. If there are no stationary points or none of them fall into the segment, then move on to the next point.
4. We calculate the values ​​of the function at selected stationary points (if any), at points at which the first derivative does not exist (if any), as well as at x=a and x=b.
5. From the obtained values ​​of the function, we select the largest and smallest - they will be the required largest and smallest values ​​of the function, respectively.

Let's analyze the algorithm for solving an example to find 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 segment [-4;-1] .

Solution.

The domain of definition of a function is the entire set of real numbers, with the exception of zero, that is. Both segments fall within the definition domain.

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].

We determine stationary points 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 the stationary point, that is, for x=1, x=2 and x=4:

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

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