IN Unified State Exam assignments In mathematics, you must meet the study of a function using its derivative. Mathematical analysis- not the easiest thing in the world. But in CMMs there is nothing that a student cannot handle high school, if he put enough effort into his studies.
Let's figure out together what a derivative is and how to use it when studying a function.
Draw a coordinate axis and construct any elementary function. For example, a parabola for the function y = x 2.
You can see for yourself that in some areas the function decreases, in others it increases. That is, it changes. This dynamics, in other words, the speed at which the function changes, reflects derivative(y" = f’(x)).
For example, mark a point on the X axis in your drawing, let our point be under number 1 - this is x 1, and on number 2 it will be x 2. Further we will operate with such concepts as argument increment – ∆х and function increment – ∆у. What it is? ∆х shows how the function changes along the X-axis, ∆у reflects the change in the function along the Y-axis.
Suppose we move along the graph from point x 1 to point x 2. Moving to the right along the X axis reflects an increment in the argument ∆x, and the resulting movement upward along the Y axis is an increment in the function ∆y. We can combine both quantities in the inequality ∆у/∆х > 0, since the increments are positive - after all, we are moving upward along an increasing graph, “in the direction of motion.”
We took two points quite far apart from each other. But in general, we can select ∆х for any point on the selected segment to obtain ∆у > 0. And on any segment where the function decreases, we can select such an increment in the argument at which ∆у< 0 и ∆у/∆х < 0.
The smaller the distance we consider, the more accurately we will describe the rate of change of the function. Not all graphs are as simple as this one. Therefore, they say that the increment of the argument tends to zero (∆x → 0), i.e. to its minimum value.
The following inequality is also possible: ∆у/∆х = 0 at the highest and lowest points of the graph. In our case, it falls on the origin of coordinates.
The inequality ∆у/∆х we wrote down reflects the essence of the derivative – we're talking about on the limit of the ratio of the increment of a function to the increment of an argument.
We started by choosing the point from which our function increment “starts”. In other words, we determined the increment of the function at the point x 1.
This means that the derivative of a function at point x 1 is the limit of the increment of the function ∆у to the increment of the argument ∆x at this point, despite the fact that ∆x → 0.
You can write this down like this: f"(x 1) = lim x→0 f (x 1 + ∆x) – f(x 1) / ∆x = lim x→0 ∆у/∆x. You can also draw a tangent to the graph at point x 1, then the derivative can be expressed through the tangent of its angle of inclination to the graph: f"(x 1) = lim x→0 ∆у/∆x = tgφ.
If the limit has boundaries (i.e. it is finite), perhaps differentiate function at a point. This will also mean that the function is continuous at this point. ∆х → 0, but ∆х ≠ 0. By the way, just because a function is continuous, it does not follow that this function can necessarily be differentiated.
If you are interested in how this can happen, I suggest you find a corresponding example yourself - not everything is ready to be received on a platter. Moreover, you don’t need to know this for the Unified State Examination tasks. And even, I’ll say something blasphemous, you may not understand what a derivative is. The main thing is to learn to find it.
Now we talked about the derivative at point x 1, but in a similar way we can perform all the same manipulations with any other point, so we have the right to write the formula for the derivative of the function as follows: f"(x) = lim x→0 f (x+ ∆x ) – f(x) / ∆х = lim x→0 ∆у/∆х. Or otherwise y" = f"(x), which occurs, is “derived” from the function y = f(x).
Here are a few derivatives as an example, you will find more of them in the table of derivatives, and some are recommended to be memorized over time:
To differentiate means to highlight certain characteristics, in the case of a function - the rate of its change, we have already talked about this. Those. calculate the derivative.
To calculate the derivative (differentiation) of a wide variety of functions, there are certain general rules. Now we will briefly recall them, using an article by Alexander Emelin from an excellent website dedicated to higher mathematics mathprofi.ru.
Y = 3cos x, y’ = (3 cos x)’ = 3 (cos x)’ = 3(-sin x) = -3sin x;
Y = 6 + x + 3x 2 – sin x – 2 3 √x + 1/x 2 – 11ctg x, y' = (6 + x + 3x 2 – sinx – 2 3 √x + 1/x 2 – 11ctg x )' = (6)' + (x)' + 3(x 2)' – (sin x)' – 2(x 1/3)'+ (x -2)' – 11(ctgx)' = 0 + 1 + 3*2x – cos x – 2*1/3x -2/3 + (-2)x -3 – 11(-1/sin 2 x) = 1 + 6x – cos x – 2/3 3 √x 2 – 2/x 3 + 11/sin 2 x;
Y = x 3 arcsin x, y' = (x 3 arcsin x)' = (x 3)' * arcsin x + x 3 * (arcsin x)'= 3x 2 arcsin x + x 3 * 1/√1 – x 2 = 3x 2 arcsin x + x 3 /√1 – x 2 ;
Y = 2(3x – 4)/ x 2 + 1, y' = (2(3x – 4)/ x 2 + 1)' = 2 (3x – 4/ x 2 +1)' = 2 * ((3x – 4)'* (x 2 + 1) – (3x – 4) * (x 2 + 1)'/(x 2 + 1) 2) = 2 (3(x 2 + 1) - (3x – 4) * 2x/ (x 2 + 1) 2) = 2 (-3x 2 + 8x + 3)/ (x 2 + 1) 2 ;
So, we’ve sorted out the saying, let’s begin the fairy tale itself. In Part B of CIMs in mathematics, you are guaranteed to come across one or even several problems involving the study of a function using the derivative. For example, you may need to examine a function for extrema, determine its monotonicity, etc.
Using the derivative you can determine:
The complexity of such tasks depends primarily on which function you meet according to the condition. But the general algorithm of actions will remain unchanged for you in any case. So let's look at everything in order.
Monotonicity of the function. Simply put, identifying areas where the function remains unchanged, i.e. "monotonous". And the function changes at critical points, but more on that below.
Procedure:
A function increases if a larger value of the function corresponds to higher value arguments: x 2 > x 1 and f(x 2) > f(x 1) on the selected interval. The graph moves from bottom to top.
A function decreases if a smaller value of the function corresponds to a larger value of the argument: x 2 > x 1 and f(x 2)< f(х 1) на выбранном интервале. График движется сверху вниз.
Since the function increases and decreases within the interval, it can be called strictly monotonic. And the study of a function for monotonicity suggests that we are talking about intervals of strict monotonicity.
The function may also not decrease on the interval: f(x 2) ≥ f(x 1) – a non-decreasing function. And similarly, do not increase on the interval: f(x 2) ≤ f(x 1) is a non-increasing function.
Sufficient conditions for the monotonicity of a function:
Critical point called the one in which the derivative is equal to zero or its value does not exist. It may at the same time be an extremum point, but it may not be one. But more on that later.
Extrema of a function. Those. such values of a variable at which the function reaches its maximum and minimum values.
Procedure:
Necessary condition for an extremum:
As mentioned above, the extremum point may not coincide with the critical point. For example, for the function y = x 3 (Fig. 1), y =│x│ (Fig. 2), y = 3 √x the extremum point is absent at the critical point.
Sufficient conditions for an extremum:
If, when passing through the point x 0, the sign of the derivative changes from “+” to “-”, then at this point the function reaches its maximum: f"(x) > 0 at x< х 0 и f"(х) < 0 при х >x 0 .
If, when passing through the point x 0, the sign of the derivative changes from “-” to “+”, then at this point the function reaches its minimum: f"(x)< 0 при х < х 0 и f"(х) >0 for x > x 0.
On the graph, extremum points reflect values along the X-axis, and extrema – values along the Y-axis. They are also called dots local extremum And local extremes. But right now, knowing the differences between local and global You won’t need extreme values, so we won’t dwell on this.
The maximum and minimum of a function are not identical concepts with its largest and smallest value. About what this is, below.
The largest and smallest values of a function that is continuous on an interval. We consider the function on the selected interval. If a function within its limits is continuous, then its greatest and smallest value on a segment fall either on the critical points that belong to it or on the points at its ends.
Procedure:
Why do we need to study a function using its derivative? Then to better understand what her schedule looks like. Yes, now in textbooks you have ready-made graphs for well-studied elementary functions. But in real “field” conditions, the situation is often exactly the opposite: an unfamiliar function and a graph that does not yet exist. And not all functions are as simple as in school textbooks. It is impossible to imagine their graphs with just the power of imagination.
Mathematical analysis tools allow you to thoroughly explore an unknown function. Without examining in detail all the characteristics of a function and its derivative, it is impossible to construct a correct graph. That is why in the school mathematics course such attention is paid to relevant tasks. And that's why they were put to the exam.
Part B tasks are worth quite high points. Therefore, pay due attention to training in determining the derivative and studying the function with its help. This article was created as a useful summary for self-study. Which contains key definitions, retold whenever possible in simple language. And it summarizes the steps you should take when researching a function.
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MOU average comprehensive school № 18.
"Exploring a function using its derivative."
Abstract on mathematics for Science Day.
Performed:
student of 11 “B” class
Bokareva Irina Nikolaevna
Supervisor:
mathematic teacher
Batyukova Galina Viktorovna.
Smolensk 2005
Introduction. 3
Chapter I. Development of the concept of function. 4
Chapter II. Basic properties of the function. 7
2.1. Definition of a function and graph of a function. Scope and
function range. Function zeros. 7
2.2. Types of functions (even, odd, general view, periodic
functions). 8
2.3. Increasing and decreasing functions. Extremes. 10
Chapter III. Research of functions. 12
3.1. General scheme Functional studies. 12
3.2. A sign of increasing and decreasing functions. 12
3.3. Critical points of a function, maxima and minima. 13
3.4. The largest and smallest values of a function. 14
Chapter IV. Examples of applying the derivative to the study of a function. 15
Conclusion. 22
References 23
Introduction.
Studying the properties of a function and plotting its graph is one of the most wonderful applications of derivatives. This method of studying function has been repeatedly subjected to careful analysis. The main reason is that in applications of mathematics it was necessary to deal with more and more complex functions that appeared in the study of new phenomena. Exceptions to the rules developed by mathematics appeared, cases appeared when the rules created were not suitable at all, functions appeared that did not have a derivative at any point.
The purpose of studying the course of algebra and beginnings of analysis in grades 10-11 is the systematic study of functions, disclosure of the applied meaning common methods mathematics related to the study of functions.
Having chosen the topic of the essay “Study of a function using the derivative,” I set the following tasks:
Systematize your knowledge about function as the most important mathematical model;
Improve your ability to use differential calculus to study elementary functions.
The development of functional concepts in the course of studying algebra and the beginnings of analysis at the senior level of education helps high school students to obtain a visual understanding of the continuity and discontinuities of functions, learn about the continuity of any elementary function in the field of its application, learn to construct their graphs and generalize information about the basic elementary functions and realize their role in the study of phenomena of reality, in human practice.
Working on the content of the topic “Studying functions using derivatives” will increase the level of my mathematical training and will allow me to solve problems more high complexity compared to the required course.
Chapter I. Development of the concept of function.
Fundamentally new part The algebra course is devoted to the study of the principles of analysis. Mathematical analysis is a branch of mathematics that took shape in the 18th century and includes two main parts: differential and integral calculus. Analysis arose thanks to the efforts of many mathematicians and played a huge role in the development of natural science - a powerful, fairly universal method for studying functions that arise when solving various applied problems appeared. Introduction to basic concepts and methods of analysis is one of the most important goals of the course.
Since the 18th century, one of the most important concepts has been the concept of function. It played and still plays a big role in understanding the real world.
The necessary prerequisites for the emergence of the concept of function were created when analytical geometry arose, characterized by the active involvement of algebra in solving geometric problems.
The idea of functional dependence arose in ancient times. It is already contained in the first mathematically expressed relationships between quantities, in the first rules for operations with numbers, in the first formulas for finding the area and volume of certain figures and geometric bodies.
However, the explicit and completely conscious use of the concept of function and the systematic study of functional dependence originates in the 17th century in connection with the penetration of the idea of variables into mathematics.
There was no clear definition of the concept of function in the 17th century, but Descartes paved the way for the first such definition. Gradually, the concept of a function began to be identified with the concept of an analytical expression - a formula.
An explicit definition of a function was first given in 1718 by Johann Bernoulli: “A function of a variable quantity is a quantity formed in any way from this variable quantity and constants.”
Leonhard Euler in his “Introduction to the Analysis of Infinites” (1748) adheres to the definition of his teacher I. Bernoulli, somewhat clarifying it. True, he did not always adhere to the above definition. Euler gives more broad meaning function, understanding it as a curve drawn by the “free pull of the hand.”
In Differential Calculus, published in 1755, Euler gives general definition functions: “When some quantities depend on others in such a way that when the latter change, they themselves are subject to change, then the first are called functions of the second.”
A great contribution to the resolution of disputes was made by Jean Baptiste Joseph Fourier, who was the first to give examples of functions that are specified in different areas by various analytical expressions.
In the second half of the 19th century, the concept of function was formulated as follows: if each element X sets A a certain element of the set is assigned IN, then we say that the function y=f(x) is given on set A, or that set A is mapped to set B.
The general concept of a function is applicable, of course, not only to quantities and numbers, but also to other mathematical objects, for example, to geometric figures.
This general definition of function was already formed in the 18th century and the first half of the 19th century. But from the very beginning of the 20th century, this definition began to raise some doubts among some mathematicians.
Dirac introduced the so-called delta function, which went far beyond the classical definition of a function.
Sergei Lvovich Sobolev was the first to consider a special case of a generalized function, including the delta function, and applied the created theory to solve a number of problems in mathematical physics.
An important contribution to the development of the theory of generalized functions was made by students and followers of L. Schwartz - I.M. Gelfand, G.E. Shilov and others.
A brief review of the development of the concept of function leads to the idea that evolution is far from over and will probably never end, just as the evolution of mathematics as a whole will never end.
Chapter II. Basic properties of the function.
2.1. Definition of a function and graph of a function. The domain of definition and the range of values of a function. Function zeros.
The ability to depict geometrically functional dependencies specified by formulas is especially important for successfully mastering a course in higher mathematics.
As is known, functional dependence is the law according to which each value of the quantity x from a certain set of numbers, called the domain of definition of the function, is associated with one well-defined value of the quantity y; the set of values that the dependent variable y takes is called the area of change of the function.
The independent variable x is also called the argument of the function. The number y corresponding to the number x is called the value of the function f at point x and is denoted f(x).
The function can be specified in three ways: analytical, tabular, graphical.
Analytical- using formulas.
Tabular– using tables where you can specify function values, but only for a finite set of argument values.
Graphic The method of specifying a function is very convenient: it makes it possible to visualize the properties of the function.
The graph of a function f is the set of all points (x;y) of the coordinate plane, where y=f(x), and x “runs through” the entire domain of definition of the function f.
Example 1 . Find the domain of the function y=lg (2x-3)
Answer: D(y)=(1.5; +∞).
One of the concepts for studying a function is the zeros of the function.
The zeros of a function are the points at which the function takes the value zero.
Example 2. Find the zeros of the function y=x 2 -5x.
A-priory:
Answer: the zeros of the function are the points x=0 and x=5.
Example 3. Find the zeros of the function y=4x-8
A-priory:
y=0, then
Answer: the zeros of this function are the point x=2.
2.2. Types of functions (even, odd, general form, periodic functions).
Let us consider functions whose domains of definition are symmetrical about the origin, that is, for any x from the domain of definition, the number (-x) also belongs to the domain of definition. These functions include even and odd.
Definition: A function f is called even if for any x from its domain f(-x)=f(x).
Schedule even function symmetrical about the ordinate axis.
Example 4. Determine the type of function y=2cos2x.
y=2cos2x, D(y)=R
y(-x)=2cos2(-x)=-2cos2x=2cos2x=y(x) – even.
Example 5. Determine the type of function y=x 4 -2x 2 +2.
y=x 4 -2x 2 +2, D(y)=R.
y(-x)=(-x) 4 -2(-x) 2 +2=x 4 -2x 2 +2=y(x) – even.
Definition: A function f is called odd if for any x from its domain f(-x)=-f(x).
Schedule odd function symmetrical about the origin.
Example 6. Determine the type of function y=2sin2x.
y=2sin2x, D(y)=R
y(-x)=2sin2(-x)=-2sin2x=-y(x) – odd.
Example 7. Determine the type of function y=3x+1/3x.
y(-x)=3(-x)+1/3(-x)=-3x-1/3x=-(3x+1/3x)=-y(x) – odd.
Example 4. Example 5.
Definition: A function f is called periodic with period T≠ 0 if for any x from the domain of definition the values of this function at points x, x-T and x+T are equal, that is, f(x+T)=f(x)=f(x-T ).
Example 8. Determine the period of the function y=cos2x.
cos2x=cos2(x+T)=cos(2x+2T), where 2T=2π, i.e. T=π.
To construct a graph of a periodic function with a period T, it is enough to construct it on a segment of length T and then transfer the resulting graph in parallel to distances nT to the right and left along the Ox axis.
Example 9. Draw a graph of the periodic function f(x)=sin2x.
sin2x=sin2(x+T)=sin(2x+2T), where 2Т=2π, i.e. T=π.
2.3. Increasing and decreasing functions. Extremes.
Also, the properties of a function include increasing and decreasing function, extrema.
The function f increases on the set P if for any x 1 and x 2 from the set P such that x 2 >x 1 the inequality f(x 2)>f(x 1) is satisfied.
The function f decreases on the set P if for any x 1 and x 2 from the set P, such that x 2 >x 1, the inequality f(x 2) is satisfied In other words, a function f is said to be increasing on the set P if a larger value of the argument from this set corresponds to a larger value of the function. A function f is said to be decreasing on the set P if a larger value of the argument corresponds to a smaller value of the function. When plotting graphs of specific functions, it is useful to first find the minimum (x min) and maximum (x max) points. A point x 0 is called a maximum point of a function f if for all x from some neighborhood of x 0 the inequality f(x) ≤f(x 0) holds. A point x 0 is called a minimum point of a function f if for all x from some neighborhood of x 0 the inequality f(x)≥ f(x 0) holds. The minimum and maximum points are usually called extremum points. Example 10.
Find the extremum points, extrema of the function y=x 2 +2x, and indicate the intervals of increase and decrease of the function. y=x 2 +2x, D(y)=R y’=(x 2 +2x)’=2x+2 y’=0, i.e. 2x+2=0 Let us examine the sign of the derivative to the right and left of the extreme point. x=-2, y’=-4+2<0 x=0, y’=0+2>0 Since the derivative changes its sign from “-” to “+”, then x = -1, this is the minimum point of the function. Since the function is continuous at the point x=-1, the function increases by [-1;+∞] and decreases by [-∞;-1]. Extremum points: x min = -1 Function extrema: y min =y(-1)=1-2= -1 Chapter III. Research of functions. 3.1. General scheme for studying functions. When examining a function, you need to know the general research scheme: 1) D(y) – domain of definition (range of change of variable x) 2) E(y) – area of x value (area of change of variable y) 3) Type of function: even, odd, periodic or general function. 4) Points of intersection of the function graph with the Ohi O axes (if possible). 5) Intervals of constancy of signs: a) the function takes a positive value: f(x)>0 b) negative value: f(x)<0. 6) Intervals of monotonicity of the function: a) increase; b) decreasing; c) constancy (f=const). 7) Extremum points (minimum and maximum points) 8) Function extrema (function value at minimum and maximum points) 9) Additional points. They can be taken in order to more accurately plot the function graph. It should be noted that the extrema of the function f do not always coincide with the largest and smallest values of the function. 3.2. A sign of increasing and decreasing functions. If you build a graph of a function using some randomly selected points, connecting them with a smooth line, then even with a very large number of randomly selected points, it may turn out that the graph constructed in this way will be very different from the graph of the given function. If you use the derivative when studying a function and find the so-called “reference” points, i.e. break points, maximum and minimum points, intervals of monotonicity of a function, then even with a small number of such “reference” points we will get a correct idea of the graph of the function. Before turning to examples, I will give the necessary definitions and theorems. Determination of monotonicity of a function on an interval
A function y=f(x) is said to be increasing on an interval if for any points x 1 and x 2 of this interval from the condition x 1<х 2
следует, что f(x 1) A sufficient sign of the monotonicity of a function in the interval.
Theorem: if a function has a positive (negative) derivative at each point of the interval, then the function increases (decreases) on this interval. This theorem is accepted without proof in school textbooks. The geometric interpretation of the theorem is very simple if we remember that f ’(x)=tgα, α is the slope of the tangent to the graph of the function at a given point x. If, for example, f ‘ (x)>0 at all points of a certain interval, then the tangent to the graph with the abscissa axis forms acute angles, which means that as x increases, f(x) also increases. If f ‘ (x)<0, то касательная с осью абсцисс образуют тупой угол, а значит, с ростом х функция f(x) убывает. Поскольку эти рассуждения основаны лишь на наглядных геометрических представлениях, они не являются доказательством теоремы. 3.3. Critical points of a function, maxima and minima. Determining the extremum points of a function
. Let x 0 be an internal point from the domain of definition of the function f(x). Then, if there is such a δ - neighborhood ] x 0 - δ, x 0 + δ [ points x 0 such that for all x from this neighborhood the inequality f(x)≤f(x 0) (the inequality f(x)≥f (x 0)), point x 0 is called the maximum point (minimum point) of this function. The maximum and minimum points are internal points of the function's domain of definition. A necessary sign of the existence of an extremum of a differentiable function
. Fermat's theorem.
If x 0 is the extremum point of the function f(x) and at this point the derivative exists, then it is equal to zero: f ’(x 0) = 0. This theorem is not a sufficient condition for the existence of an extremum of a differentiable function: if at some point x 0 the derivative vanishes, then it does not follow from this that the function has an extremum at the point x 0. Determining critical points of a function
. The interior points of the domain of definition of a function at which its derivative is equal to zero or does not exist are called critical points of the function. Sufficient conditions for the existence of an extremum
. Theorem 1.
If the function f(x) is continuous at the point x 0, f ‘(x)>0 on the interval and f ‘(x)<0 на интервале , то х 0
является точкой максимума функции f(x). Theorem 2.
If the function f(x) is continuous at the point x 0, f ‘(x)<0 на интервале и f ‘(x)>0 on the interval , then x 0 is the minimum point of the function f(x). To find the extreme points of a function, you need to find its critical points and for each of them check whether sufficient conditions for the extremum are met. 3.4. The largest and smallest values of a function. Rules for finding the largest and smallest values of functions in the interval.
To find the largest and smallest values of a function differentiable in a certain interval, you need to find all the critical points lying inside the interval, calculate the values of the function at these points and at the ends of the interval, and select the largest and smallest from all the values of the function obtained in this way. Chapter IV. Examples of applying the derivative to the study of a function. Example 11.
Explore the function y=x 3 +6x 2 +9x and draw a graph. 2) Let's determine the type of function: y(-x)=(-x) 3 +6(-x) 2 +9(-x)=-x+6x 2 -9x function of general form. x=0 or x 2 +6x+9=0 D=0, the equation has one root. (0;0) and (-3;0) are the points of intersection with the x-axis. y’=(x 3 +6x 2 +9x)’=3x 2 +12x+9 y’=0, i.e. 3x 2 +12x+9=0 reduce by 3 D>0, the equation has 2 roots. x 1,2 =(-b±√D)/2a, x 1 =(-4+2)/2, x 2 =(-4-2)/2 0 x=-4, y’=3*16-48+9=9>0 x=-2, y’=12-24+9=-3<0 x=0, y’=0+0+9=9>0 7) Find x min and x max: 8) Find the extrema of the function: y min =y(-1)=-1+6-9=-4 y max =y(-3)=-27+54-27=0 9) Let's plot the function: 10) Additional points: y(-4)=-64+96-36=-4 Example 12.
Explore the function y=x 2 /(x-2) and plot a graph y=x 2 /(x-2)=x+2+4/(x-2) Let's find the asymptotes of the function: y=x+2 – oblique asymptote, because Let's find the domain of definition. 2) Let's determine the type of function. y(-x)=(-x) 2 /(-x-2)=x 2 /(-x-2), a function of general form. 3) Find the points of intersection with the axes. Oy: x=0, y=0 (0;0) – point of intersection with the y-axis. x=0 or x=2 (2;0) – point of intersection with the x axis 4) Find the derivative of the function: y'=(2x(x-2)-x 2)/(x-2) 2 =(2x 2 -4x-x 2)/(x-2) 2 =(x(x-4))/(x -2) 2 =(x 2 -4x)/(x-2) 2 5) Let's determine the critical points: x 2 -4x=0 x(x-4)=0 y’=0, (x 2 -4x)/(x-2) 2 =0<=> <=> (x-2) 2 ≠ 0 x≠ 2 x 2 -4x=0, and (x-2) 2 ≠ 0, i.e. x≠ 2 6) Let us designate the critical points on the coordinate line and determine the sign of the function. x=-1, y’=(1+4)/9=5/9>0 x=1, y’=(1-4)/1=-3<0 x=3, y’=(9-12)/1=-3<0 x=5, y’=(25-20)/9=5/9>0 7) Find the minimum and maximum points of the function: 8) Find the extrema of the function: y min =y(4)=16/2=8 9) Let's plot the function: 10) Additional points: y(-3)=9/-5=-1.8 y(3)=9/1=9 y(1)=1/-1=-1 y(6)=36/4=9 Example 13.
Explore the function y=(6(x-1))/(x 2 +3) and construct a graph. 1) Find the domain of definition of the function: 2) Let's determine the type of function: y(-x)=(6(-x-1))/(x 2 +3)=-(6(x+1))/(x 2 -3) is a function of general form. 3) Find the points of intersection with the axes: O y: x=0, y=(6(0-1))/(0+3)=-2, (0;-2) – point of intersection with the y axis. (6(x-1))/(x 2 +3)=0 O x: y=0,<=> 4) Find the derivative of the function: y'=(6(x-1)/(x 2 +3))'=6(x 2 +3-2x 2 +2x)/(x 2 +2) 2 =-6(x+1)(x -3)/(x 2 +3) 2 5) Let's determine the critical points: y’=0, i.e. -6(x+1)(x-3)/(x 2 +3) 2 =0 y’=0, if x 1 =-1 or x 2 =3, then x=-1 and x=3, critical points. 6) Let us denote the critical points on the coordinate line and determine the sign of the function: x=-2, y’=-6(-2+1)(-2-3)/(4+3) 2 =-30/49<0 x=0, y’=-6(0+1)(0-3)/(0+3) 2 =2>0 x=4, y’=-6(4+1)(4-3)/(16+3) 2 =-30/361<0 7) Find the minimum and maximum points: 8) Find the extrema of the function: y min =y(-1)=(6(-1-1))/(1+3)=-12/4=-3 y max =y(3)=(6(3-1))/(9+3)=12/12=1 9) Let's plot the function: 10) Additional points: y(-3)=(6(-3-1))/(9+3)=-24/12=-2 y(6)=(6(6-1))/(36+3)=30/39=10/13≈ 0.77 Example 14.
Explore the function y=xlnx and plot it: 1) Find the domain of definition of the function: D(y)=R + (positive values only) 2) Let's determine the type of function: y(-x)=-xlnx - of general form. 3) Find the points of intersection with the axes: O y, but x≠ 0, which means there are no points of intersection with the y axis. O x: y=0, that is xlnx=0 x=0 or lnx=0 (1;0) – point of intersection with the x axis 4) Find the derivative of the function: y’=x’ ln x + x(ln x)’=ln x +1 5) Let's determine the critical points: y’=0, that is lnx +1=0 y’=0, if x=1/e, then x=1/e is the critical point. 6) Let us denote the critical points on the coordinate line and determine the sign of the function: 1/e x=1/(2e); y’=log(2e) -1 +1=1-ln(2e)=1-ln e=-ln 2<0 x=2e; y’=ln(2e)+1=ln 2+ln e+1=ln 2+2>0 7) 1/e – minimum point of the function. 8) Find the extrema of the function: y min =y(1/e)=1/e ln e -1 =-1/e (≈ -0.4). 9) Let's plot the function: Conclusion. Many scientists and philosophers have worked on this topic. Many years ago these terms originated: function, graph, study of function, and they have still been preserved, acquiring new features and characteristics. I chose this topic because I was very interested in going through this path of research into function. It seems to me that many would be interested in learning more about the function, its properties and transformations. By completing this essay, I systematized my skills and expanded my knowledge about this topic. I would like to encourage everyone to study this topic further. Bibliography. 1. Bashmakov, M.I. Algebra and the beginning of analysis. - M.: Education, 1992. 2. Glazer, G.I. History of mathematics in school. - M.: Education, 1983. 3. Gusev, V.A. Mathematics: Reference materials. - M.: Education, 1888. 4. Dorofeev, G.V. A manual on mathematics for those entering universities. - M.: Nauka, 1974. 5. Zorin, V.V. A manual on mathematics for those entering universities. - M.: Higher School, 1980. 6. Kolmogorov A.N. Algebra and the beginnings of analysis. - M.: Education, 1993. Purpose of the lesson: testing the skills of studying functions and plotting graphs using derivatives. Theoretical part of the test. Questions
Determining the minimum and maximum points. Theoretical part of the test 1) Determination of the minimum point. If the function is defined in some neighborhood of the point X 0, then the point X 0 is called minimum point
functions
f(x), if there is a neighborhood of the point X 0 such that for all xx 0 from this neighborhood the inequality f(x)>f(x 0) is satisfied. Determination of the maximum point. If the function is defined in some neighborhood of the point X 0, then the point X 0 is called maximum point
functions
f(x), if there is a neighborhood of the point X 0 such that for all x? x 0 from this neighborhood the inequality f(x) is satisfied 2) Determination of critical points. Critical points are internal points of the domain of definition of a function at which the derivative does not exist or is equal to zero. 3) A necessary condition for X 0 to be a point extremum
: This point must be critical. 4) Algorithm for finding critical points. 1. Find the domain of definition of the function. 2. Find the derivative of the function. 3. Find the domain of definition of the derivative of a given function. (To determine whether there are points at which the derivative does not exist. If there are such points, then check whether they are internal points of the domain of definition of the function. 4. Find the points at which the derivative is equal to zero by solving the equation: f "(x)=0. Check whether the found points are internal points of the function domain. 5) Stationary points - points at which the derivative of the function is equal to zero. 6) Fermat's theorem. (Prerequisite extremum of the function.) y=f(x) is a function that is defined in a certain neighborhood of the point X 0 and has a derivative at this point. Theorem: if X 0 is the extremum point of the differentiable function f(x), then f "(x)=0. 7) Sufficient conditions for the existence of an extremum functions at a point. y=f(x) is defined on (a;c). X 0 is the critical point. If the function f is continuous at point X 0, and f "(x)>0 on the interval (a; x 0) and f "(x)<0 на
интервале (х 0 ;в), то точка х 0 является maximum point of the function f. (Simplified formulation: if at point X 0 the derivative changes sign from “+” to “_”, then X 0 there is a maximum point.) If the function f is continuous at point X 0, and f "(x)<0 на интервале (а;X 0) и f "(х)>0 on the interval (X 0 ;в), then the point x 0 is minimum point of the function f. (Simplified formulation: if at point X 0 the derivative changes sign from “_” to “+”, then X 0 is minimum point.) 8) Sufficient sign of increase, descending
functions
. If f "(x)>0 for all x from the interval (a; b), then the function increases on the interval (a; b). If f "(x)<0 для всех х из промежутка (а; в), то
функция убывает на промежутке (а; в). (If the function is continuous at the end of the interval, then it can be added to the interval of increasing (decreasing) function.) 9) Extremum points, extremum of the function. X 0 - maximum point, X 0 - minimum point are called extremum points. f(x 0) - maximum of the function, f(x 0) - the minimum of the function is called extrema of the function. 10) Algorithm for finding the extrema of a function. 1. Find the domain of definition of the function. 2. Find the derivative of the function. 3. Find critical points. 4. Let us determine the sign of the derivative on each of the intervals into which the critical points divide the domain of definition. 5. Let us find the extremum points, taking into account the nature of the change in the sign of the derivative. 6. Let's find the extrema of the functions. 11) Algorithm for finding the largest and the smallest values of a function on a segment.
1. Find the values of the function at the ends of the segment [a; V]. 2. Find the values of the function at those critical points that belong to the interval (a; b). 3. From the found values, select the largest and smallest. Practical part of the test “Study of functions using derivatives. The largest and smallest values of functions on a segment” a) critical points of functions, b) extrema of functions c) the largest and smallest values of functions on the specified interval d) build a graph. In problem B15 it is proposed to examine the function specified by the formula for extrema. This is a standard calculus problem, and its difficulty varies greatly depending on the function in question: some can be solved literally orally, while others require serious thought. Before studying solution methods, you need to understand some terms from the field of mathematical analysis. So, in Problem B15 you need to find the following quantities using the derivative: In this case, global extrema are usually sought not over the entire domain of definition of the function, but only over a certain segment. It is important to understand that the global extremum and the value of the function at the extremum point do not always coincide. Let's explain this with a specific example: Task. Find the minimum point and minimum value of the function y = 2x 3 − 3x 2 − 12x + 1 on the interval [−3; 3]. First, we find the minimum point, for which we calculate the derivative: Let’s find the critical points by solving the equation y’ = 0. We get the standard quadratic equation: Let's mark these points on the coordinate line, add derivative signs and restrictions - the ends of the segment: The scale of the picture does not matter. The most important thing is to mark the points in the correct sequence. From a school mathematics course we know that at the minimum point the derivative changes sign from minus to plus. The count always goes from left to right - in the direction of the positive semi-axis. Therefore, there is only one minimum point: x = 2. Now let’s find the minimum value of the function on the interval [−3; 3]. It is achieved either at the minimum point (then it becomes the global minimum point) or at the end of the segment. Note that on the interval (2; 3) the derivative is positive everywhere, which means y(3) > y(2), so the right end of the segment can be ignored. The only points left are x = −3 (the left end of the segment) and x = 2 (the minimum point). We have: So, the smallest value of the function is achieved at the end of the segment and is equal to −44. Answer: x min = 2; y min = −44 From the above reasoning follows an important fact that many people forget. The function takes its maximum (minimum) value not necessarily at the extremum point. Sometimes this value is reached at the end of the segment, and the derivative there does not have to be equal to zero. If in problem B15 you need to find the maximum or minimum value of the function f(x) on the interval, perform the following steps: A short explanation about crossing out roots when they coincide with the ends of a segment. They can also be crossed out, since in the fourth step the ends of the segment are still substituted into the function - even if the equation f’(x) = 0 had no solutions. Task. Find highest value functions y = x 3 + 3x 2 − 9x − 7 on the interval [−5; 0]. First, let’s find the derivative: y’ = (x 3 + 3x 2 − 9x − 7)’ = 3x 2 + 6x − 9. Then we solve the equation: y’ = 0 ⇒ 3x 2 + 6x − 9 = 0 ⇒ ... ⇒ x = −3; x = 1. We cross out the root x = 1, because it does not belong to the segment [−5; 0]. It remains to calculate the value of the function at the ends of the segment and at the point x = −3: Obviously, the largest value is 20 - it is achieved at the point x = −3. Now consider the case when you need to find the maximum or minimum point of the function f(x) on the segment. If the segment is not specified, the function is considered in its domain of definition. In any case, the solution is as follows: The thoughtful reader will probably notice that for some functions this algorithm does not work. Indeed, there is a whole class of functions for which finding extremum points requires more complex calculations. However, such functions are not found in the Unified State Examination in mathematics. Pay careful attention to the placement of signs between the points x 1, x 2, ..., x n. Remember: when passing through a root of even multiplicity, the sign of the derivative does not change. When looking for extreme points, the signs are always viewed from left to right, i.e. in the direction of the number axis. Task. Find the maximum point of the function on the segment [−8; 8]. Let's find the derivative: Since this is a fractional rational function, we equate the derivative and its denominator to zero: Let's mark the points x = −5, x = 0 and x = 5 on the coordinate line, place signs and boundaries: Obviously, there is only one point left inside the segment x = −5, at which the sign of the derivative changes from plus to minus. This is the maximum point. Let us once again explain how the extremum points differ from the extrema themselves. Extremum points are the values of variables at which the function takes on the largest or smallest value. Extrema are the values of the functions themselves, maximum or minimum in some of their neighborhoods. In addition to the usual polynomials and fractional rational functions, the following types of expressions are found in Problem B15: As a rule, no problems arise with irrational functions. The remaining cases are worth considering in more detail. The main difficulty with trigonometric functions is that when solving equations, an infinite number of roots arise. For example, the equation sin x = 0 has roots x = πn, where n ∈ Z. Well, how to mark them on the coordinate line if there are infinitely many such numbers? The answer is simple: you need to substitute specific values of n. Indeed, in problems B15 with trigonometric functions there is always a constraint - a segment. Therefore, to begin with, we take n = 0, and then increase n until the corresponding root “flies” beyond the boundaries of the segment. Similarly, by decreasing n, we will very soon obtain a root that is less than the lower bound. It is easy to show that no roots, other than those obtained in the process considered, exist on the segment. Let us now consider this process using specific examples. Task. Find the maximum point of the function y = sin x − 5x·sin x − 5cos x + 1, belonging to the segment [−π/3; π/3]. We calculate the derivative: y’ = (sin x − 5x sin x − 5cos x + 1)’ = ... = cos x − 5x cos x = (1 − 5x) cos x. Then we solve the equation: y’ = 0 ⇒ (1 − 5x) cos x = 0 ⇒ ... ⇒ x = 0.2 or x = π/2 + πn, n ∈ Z. Everything is clear with the root x = 0.2, but the formula x = π/2 + πn requires additional processing. We will substitute different values of n, starting from n = 0. n = 0 ⇒ x = π/2. But π/2 > π/3, so the root x = π/2 is not included in the original segment. Also, the larger n, the larger x, so there is no point in considering n > 0. n = −1 ⇒ x = − π/2. But −π/2< −π/3 - этот корень тоже придется отбросить. А вместе с ним - и все корни для n < −1. It turns out that on the interval [−π/3; π/3] lies only with the root x = 0.2. Let's mark it along with the signs and boundaries on the coordinate line: To make sure that the derivative to the right of x = 0.2 is really negative, it is enough to substitute the value x = π/4 into y’. We will simply note that at the point x = 0.2 the derivative changes sign from plus to minus, and therefore this is the maximum point. Task. Find the largest value of the function y = 4tg x − 4x + π − 5 on the interval [−π/4; π/4]. We calculate the derivative: y’ = (4tg x − 4x + π − 5)’ = 4/cos 2x − 4. Then we solve the equation: y’ = 0 ⇒ 4/cos 2x − 4 = 0 ⇒ ... ⇒ x = πn, n ∈ Z. Let us extract the roots from this formula by substituting specific n, starting from n = 0: Of the entire variety of roots, only one remains: x = 0. Therefore, we calculate the value of the function for x = 0, x = π/4 and x = −π/4. Now note that π = 3.14...< 4, поэтому π − 5 < 4 − 5 = −1 и 2π − 9 < 8 − 9 = −1. Получается одно положительное число и два отрицательных. Мы ищем наибольшее - очевидно, это y = −1. Note that in the last problem it was possible not to compare the numbers with each other. After all, of the numbers π − 5, 1 and 2π − 9, only one can be written on the answer form. Indeed, how to write, say, the number π on a form? But no way. This important feature the first part of the Unified State Examination in mathematics, which greatly simplifies the solution of many problems. And it works not only in B15. Sometimes when studying a function, equations arise that have no roots. In this case, the task becomes even simpler, since only the ends of the segment remain to be considered. Task. Find the smallest value of the function y = 7sin x − 8x + 5 on the interval [−3π/2; 0]. First we find the derivative: y’ = (7sin x − 8x + 5)’ = 7cos x − 8. Let's try to solve the equation: y’ = 0 ⇒ 7cos x − 8 = 0 ⇒ cos x = 8/7. But the values of cos x always lie on the interval [−1; 1], and 8/7 > 1. Therefore, there are no roots. If there are no roots, then there is no need to cross out anything. Let's move on to the last step - calculate the value of the function: Since the number 12π + 12 cannot be written on the answer sheet, all that remains is y = 5. Generally speaking, an exponential function is an expression of the form y = a x, where a > 0. But in problem B15 there are only functions of the form y = e x and, in extreme cases, y = e kx + b. The reason is that the derivatives of these functions are calculated very easily: Everything else is absolutely standard. Of course, the real functions in problems B15 look more severe, but this does not change the solution scheme. Let's look at a couple of examples, highlighting only the main points of the solution - without thorough reasoning or commentary. Task. Find the smallest value of the function y = (x 2 − 5x + 5)e x − 3 on the interval [−1; 5]. Derivative: y’ = ((x 2 − 5x + 5)e x − 3)’ = ... = (x 2 − 3x)e x − 3 = x(x − 3)e x − 3 . Find the roots: y’ = 0 ⇒ x(x − 3)e x − 3 = 0 ⇒ ... ⇒ x = 0; x = 3. Both roots lie on the segment [−1; 5]. It remains to find the value of the function at all points: Of the four numbers obtained, only y = −1 can be written on the form. In addition, this is the only negative number - it will be the smallest. Task. Find the largest value of the function y = (2x − 7) e 8 − 2x on the segment. Derivative: y’ = ((2x − 7) e 8 − 2x)’ = ... = (16 − 4x) e 8 − 2x = 4(4 − x) e 8 − 2x . Find the roots: y’ = 0 ⇒ 4(4 − x) e 8 − 2x = 0 ⇒ x = 4. The root x = 4 belongs to the segment . We are looking for the function values: Obviously, only y = 1 can be the answer. By analogy with exponential functions, in problem B15 only natural logarithms are encountered, since their derivative is easily calculated: Thus, the derivative will always be a fractional rational function. All that remains is to equate this derivative and its denominator to zero, and then solve the resulting equations. To find the maximum or minimum value of a logarithmic function, remember: natural logarithm turns into a “normal” number only at points of the form e n . For example, ln 1 = ln e 0 = 0 is a logarithmic zero, and most often the solution comes down to it. In other cases, it is impossible to “remove” the sign of the logarithm. Task. Find the smallest value of the function y = x 2 − 3x + ln x on the segment. We calculate the derivative: We find the zeros of the derivative and its denominator: Of the three numbers x = 0, x = 0.5 and x = 1, only x = 1 lies inside the segment, and the number x = 0.5 is its end. We have: Of the three values obtained, only y = −2 does not contain a logarithm sign - this will be the answer. Task. Find the greatest value of the function y = ln(6x) − 6x + 4 on the segment. We calculate the derivative: We find out when the derivative or its denominator are equal to zero: We cross out the number x = 0, since it lies outside the segment. We calculate the value of the function at the ends of the segment and at the point x = 1/6: Obviously, only y = 3 can act as an answer - the remaining values contain a logarithm sign and cannot be written on the answer sheet.
-4
x≠ 2, x=2 – vertical asymptote
0 8
-3 2
1. y=(x-3) 2 (x-2).
11. y=2x 4 -x.
[-1;1]
2. y=1/3x 3 +x 2
[-4;1]
12. y=x 2 -2/x.
[-3;-0,5]
3. y=1/3x 3 -x 2 -3x
[-2;6]
13. y=1/(x 2 +1).
[-1;2]
4. y=-1/4x 4 +2x 2 +1.
[-3;3]
14. y=3x-x 3 .
[-1,5;1,5]
5. y=x 4 -8x 2 -9.
[-3;3]
15. y=2x 2 -x 4.
[-2;1,5]
6. y=(x-2)(x+1) 2.
[-1,5;1,5]
16. y=3x 2/3 -x 2.
[-8;8]
7. y=-2/3x 3 +2x-4/3.
[-1,5;1,5]
17. y=3x 1/3 -x.
[-8;8]
8. y=3x 5 -5x 4 +4.
[-1;1]
18. y=x 3 -1.5x 2 -6x+4.
[-2;3]
9. y=9x 2 -9x 3.
[-0,5;1]
19. y=(1-x)/(x 2 +3).
[-2;5]
10. y=1/3x 3 -4x.
[-3;3]
20. y= -x 4 +2x 2 +3.
[-0,5;2]
y’ = (2x 3 − 3x 2 − 12x + 1)’ = 6x 2 − 6x − 12.
y’ = 0 ⇒ 6x 2 − 6x − 12 = 0 ⇒ ... ⇒ x 1 = −1, x 2 = 2.
y(−3) = 2(−3) 3 − 3(−3) 2 − 12(−3) + 1 = −44;
y(2) = 2*2 3 − 3*2 2 − 12*2 + 1 = −19.Problem solving scheme B15
y(−5) = (−5) 3 + 4·(−5) 2 − 9·(−5) − 7 = −12;
y(−3) = (−3) 3 + 4·(−3) 2 − 9·(−3) − 7 = 20;
y(0) = 0 3 + 4 0 2 − 9 0 − 7 = −7.
y’ = 0 ⇒ x 2 − 25 = 0 ⇒ ... ⇒ x = 5; x = −5;
x 2 = 0 ⇒ x = 0 (second multiplicity root).
Trigonometric functions
n = 0 ⇒ x = 0. This root suits us.
n = 1 ⇒ x = π. But π > π/4, so the root x = π and the values n > 1 must be crossed out.
n = −1 ⇒ x = −π. But π< −π/4, поэтому x = π и n < −1 тоже вычеркиваем.
y(0) = 4tg 0 − 4 0 + π − 5 = π − 5;
y(π/4) = 4tg (π/4) − 4 π/4 + π − 5 = −1;
y(π/4) = 4tg (−π/4) − 4 (−π/4) + π − 5 = ... = 2π − 9.
y(−3π/2) = 7sin (−3π/2) − 8·(−3π/2) + 5 = ... = 12π + 12;
y(0) = 7sin 0 − 8 0 + 5 = 5.Exponential functions
y(−1) = ((−1) 2 − 5·(−1) + 5)e − 1 − 3 = ... = 11·e −4 ;
y(0) = (0 2 − 5 0 + 5)e 0 − 3 = ... = 5 e −3 ;
y(3) = (3 2 − 5 3 + 5)e 3 − 3 = ... = −1;
y(5) = (5 2 − 5 5 + 5)e 5 − 3 = ... = 5 e 2 .
y(0) = (2 0 − 7)e 8 − 2 0 = ... = −7 e 8 ;
y(4) = (2 4 − 7)e 8 − 2 4 = ... = 1;
y(6) = (2 6 − 7)e 8 − 2 6 = ... = 5 e −4 .Logarithmic functions
y’ = 0 ⇒ 2x 2 − 3x + 1 = 0 ⇒ ... ⇒ x = 0.5; x = 1;
x = 0 - there is nothing to decide here.
y(0.5) = 0.5 2 − 3 0.5 + ln 0.5 = ln 0.5 − 1.25;
y(1) = 1 2 − 3 1 + ln 1 = −2;
y(5) = 5 2 − 3 5 + ln 5 = 10 + ln 5.
y’ = 0 ⇒ 1 − 6x = 0 ⇒ x = 1/6;
x = 0 - already decided.
y(0.1) = ln(6 0.1) − 6 0.1 + 4 = ln 0.6 + 3.4;
y(1/6) = ln(6 1/6) − 6 1/6 + 4 = ln 1 + 3 = 3;
y(3) = ln(6 3) − 6 3 + 4 = ln 18 − 14.
