Today we will talk about the study of functions. It is important to note that mathematics is arranged in the same way as ordinary house: first the foundation is laid, and then bricks are laid out layer by layer. The role of the foundation in mathematics is played by a function (correspondence between two sets). After introducing the concept of a function, they begin to study it as an object in the same way as it was done with numbers.
In fact, in life we also often use not only objects, but also correspondences between them, relations between objects. An example is books about love (love is a relationship between people).
After the study of functions in mathematics, one begins to study sets of functions, then spaces of functions, and so on. But today we will talk about the primary analysis of the function.
What is a function? A function is a correspondence between sets. In this lesson, we will talk about numerical functions, that is, about correspondences between numerical sets. We will also talk about the local property of the function (the behavior of the function at this particular point) and the global property (the property associated with the entire scope of the function). The derivative is a description of local properties of functions, and the integral is a description of global ones.
For example, there are two different functions, but at a point their graphs coincide (see Fig. 1). But what is the difference between the behavior of functions in the vicinity of this point? This will be discussed.
Rice. 1. Intersection of graphs of two different functions
From the graph of a function, you can easily determine its properties: monotonicity (function increasing or decreasing), parity (odd) and periodicity (see Fig. 2).
Rice. 2. Feature specifications
All of these characteristics are mathematical. But the derivative is often used in life. Most often, when we describe a process using a graph, we are interested in the dynamics of this process, that is, not the value of the function at a particular point, but how the function will behave in the future (will it increase or decrease?). For example, when we want to analyze price growth or compare prices over different periods of time (the absolute values could change, but the dynamics remained the same) (see Fig. 3).
Rice. 3. Dynamics of gold prices
The derivative helps to figure out how the function will behave in the neighborhood of a given point.
It is worth clarifying that in school, most often, the derivative of a function is sought over the entire domain of definition. This is due to the fact that the features under investigation are "good", that is, their behavior is predictable on the entire axis. But in general the derivative is a local characteristic of a function.
For example, when viewing photos with different shutter speeds, there may be several options:
- cars are standing and people are each in their place (see Fig. 4);
- a blurry picture, you can see who is going where (see Fig. 5).
Rice. 4. Photo with exposure to
Rice. 5. Photo with exposure to
The second option is visual illustration derivative (blurring the picture).
At the point, the function takes on a specific value, and it is practically impossible to draw any conclusions about its behavior from it. And if we consider the neighborhood of this point, then we can already say which side it is smaller (which one is larger) and conclude whether it increases or decreases. That is, when the shutter speed is short, we see the value of the function at a point, and when we consider the frame delay, we can already analyze the behavior of the function (see Fig. 6).
Rice. 6. Analogy between derivative and photography
V Everyday life we often analyze a situation like the analysis of functions in mathematics. For example, when we say that it is getting warmer (cooler) outside, we do not indicate the specific temperature at the moment, but we mean that the temperature will soon rise (decrease). This is similar to calculating the derivative (see Fig. 7).
Rice. 7. Temperature change analysis
Let's introduce precise definition derivative.
Derivative functionat the point the limit is called the ratio of the increment of the function at this point to the increment of the argument (provided that this limit exists):
Since we want to introduce such a concept as the rate of change of a function (the main word is speed), then we can draw a parallel with physics. Instantaneous speed - vector physical quantity, equal to the ratio of the movement to the time interval during which this movement occurred, if the time interval tends to zero:
Instantaneous speed, m/s; - body displacement, m (at ); - tending to zero time interval, s.
But it is important to clarify that when we talked about temperature, we indicated only the qualitative characteristics of the process, but did not talk about the rate of temperature change. The derivative takes into account the rate of change of the function. Features can grow in different ways. For example, the parabola () increases faster than the logarithm () (see Fig. 8).
Rice. 8. The rate of increase of graphs of functions and
It is to compare the rate of increase (decrease) of the function that we introduce a specific characteristic of the function - the derivative. Drawing an analogy between the derivative and the speed of movement of an object (speed is the ratio of the distance traveled to time, or the change in coordinates per unit time), we can say that in the limit, the derivative is the ratio of the change in the function (that is, the path that the point has traveled , if it moved along the graph of the function) to the increment of the argument (the time during which the movement was performed) (see Fig. 9). This is the mechanical (physical) meaning of the derivative.
Rice. 9. Analogy between speed and derivative
The derivative is a local property of a function. It is important to distinguish between the calculation of the derivative over the entire domain of definition and in a specific area, because the function on one interval could be quadratic, on the other - linear, and so on. But this is all one function, and at different points such a function will have different meanings derivative.
For most functions given analytically (by a specific formula), we have a table of derivatives (see Fig. 10). This is an analogue of the multiplication table, that is, there are basic functions for which the derivatives have already been calculated (it can be proved that they have exactly this form), and then there are some rules (see Fig. 11) (analogues of multiplication or division in a column), with which can be used to calculate derivatives complex functions, derivative works, and so on. Thus, for almost all functions expressed in terms of functions known to us, we can describe the behavior of the function over the entire domain of definition.
Rice. 10. Table of derivatives
Rice. 11. Differentiation rules
But still, the definition of the derivative, which we gave earlier, is pointwise. To generalize the derivative at a point to the entire domain of the function, it is necessary to prove that at each point the value of the derivative will coincide with the values of the same function.
If we imagine a function that is not written analytically, then in the neighborhood of each point we can represent it as a linear function. The derivative of a linear function in a neighborhood of some point is easy to calculate. If we represent a function linearly, then it coincides with its tangent (see Fig. 12).
Rice. 12. Function representation at each point as a linear function
From right triangle we know that the tangent is equal to the ratio of the opposite leg to the adjacent one. Therefore, the geometric meaning of the derivative is that the derivative is the tangent of the slope of the tangent at this point (see Fig. 13).
Rice. 13. The geometric meaning of the derivative
Speaking about the derivative as about the speed, we can say that if the function is decreasing, then its derivative is negative, and vice versa, if the function is increasing, then its derivative is positive. On the other hand, we have defined the derivative as the tangent of the slope of the tangent. This is also easy to explain. If the function is increasing, then the tangent forms an acute angle, and the tangent of the acute angle is positive. Therefore, the derivative is positive. As you can see, the physical and geometric meaning of the derivative coincided.
Acceleration is the rate of change of speed (that is, the derivative of speed). On the other hand, speed is the derivative of displacement. It turns out that acceleration is the second derivative (derivative of the derivative) of displacement (see Fig. 14).
Rice. 14. Application of the derivative in physics
A derivative is a means of studying the properties of a function.
The derivative is used to solve optimization problems. There is an explanation for this. Since the derivative shows the growth of the function, it can be used to find the local maxima and minima of the function. Knowing that the function increased in one section, and then it began to decrease, we assume that there is a local maximum at some point. Similarly, if the function was decreasing and then began to increase, at some point there is local minimum(see fig. 15).
Rice. 15. Local minima and maxima of a function
In practice, this can be used to find, for example, the maximum profit under given conditions. To do this, you need to find a point at which there will be a local maximum. If we need to define minimum costs, then, accordingly, it is necessary to determine the point at which the local minimum is located (see Fig. 16).
Rice. 16. Finding the maximum profit and minimum cost
The school solves many optimization problems. Let's consider one of them.
What should be a rectangular fence of fixed length so that it encloses the maximum area (see Fig. 17)?
Rice. 17. Optimization problem
It turns out that the fence should be square.
There are a lot of such tasks, when one parameter is fixed, and the second one needs to be optimized. The parameter that is fixed is our task data (for example, the material for the fence). And there is a parameter that we want to get the minimum or maximum (for example, the maximum area, the minimum size). That is, a pair of "resource - effect" is formed. There is some resource that is initially set, and some effect that we want to get.
Now let's move on to the global properties of the function. Consider the simplest case of an integral. Let's take a series of numbers: . A series is also a function (of a natural argument), each number has its own serial number and meaning. 
.
Let's write the formula for finding the sum of this series:
The sum up to some specific value will be the value of the integral.
For example, for:
That is, the integral is actually the sum (in this case, the sum of the function values).
Most students associate the integral with area. Let's try to connect the example with the sum of the series and the area. Let's rewrite this series as a linear function: .
Then the sum of this series will be the sum of the areas of the parts under the graph (in this case, trapezoids) (see Fig. 18).
Rice. 18. Area under the graph of a function
The sum of the areas is equal to the area of the sum (if the parts into which the figure is divided do not intersect). So the integral is the area under the graph of the function. Thus, having found the integral, we can find the area of some part of the plane. For example, you can find the area under the graph.
If we want to strictly introduce the definition of the integral in terms of the area of the figure under the function, then we need to break the figure itself into very small pieces. It is not always as convenient to calculate the area as in the case of a linear function. Let's take a function for example. If we linearly approximate the function (as we proposed to do in the case of the derivative), then, just as in the previous example, we will get a partition of the entire area into the sum of the areas of trapezoids (see Fig. 19).
Then, in the limit, this is the integral, that is, the area under the graph of the function.
Rice. 19. Area under the graph of a function
But how to calculate this area (integral)? For known functions, there is a table of integrals (similar to a table of derivatives). But in the general case, we approximate the function by segments and calculate the sum of the areas of the trapezoids under these segments. Reducing the segments, in the limit we obtain the value of the integral.
In contrast to the derivative, when a "good" derivative is always obtained for a "good" function, this is not the case in the case of an integral. For example, for such a simple function as we cannot calculate the integral and present it in the form of analytical functions (see Fig. 20).
Calculating the integral is not an easy task, and therefore the existence of such a simple Newton-Leibniz formula (see Fig. 20), which allows us to quickly calculate the value of the integral, if we know its form, greatly facilitates the calculations. Otherwise, it would be difficult to calculate the limiting area every time.
Rice. 20. Newton-Leibniz formula for calculating integrals
Therefore, the main calculation methods are:
- table of integrals for those functions that we can calculate (see Fig. 21);
- integral properties that allow you to calculate different combinations of table functions (see Fig. 22);
- Newton-Leibniz formula (if we calculate the value at the extreme right point and subtract the value at the extreme left point, we get the area) (see Fig. 20).
Rice. 21. Table of integrals
Rice. 22. Properties of a definite integral
At school, the Newton-Leibniz formula is not derived, although this is not difficult to do if you define the integral as the area under the graph.
More about the derivation of the Newton-Leibniz formula:
To better understand the difference between local and global properties of a function, consider the example of target shooting. If you take several shots around (none hit the center) and calculate the average, you get practically (see Fig. 23). Although in fact the shooter could hit all the time above or below the target, the average would still turn out to be close to .
Rice. 23. Target shooting
We can give an example from physics - the center of gravity. The same mass with the same center of gravity can be distributed in completely different ways (see Fig. 24).
Rice. 24. Variants of mass distribution with the same center of gravity
As another example, one can average temperature by hospital. If someone has a temperature, and someone has it, then on average it turns out and it seems that patients are not so sick.
If we talk about the connection between the derivative (local characteristic) and the integral (global characteristic), then it is intuitively clear that these are mutually inverse concepts. In fact, it is. If we take the derivative of the integral or the integral of the derivative, we get the original function. To explain this, consider the motion of a body. We already know that speed is the derivative of displacement. Let's try to perform the reverse operation. To do this, we express the movement in terms of speed and time:
And if we look at the graph (speed changes linearly), we will see that the path is the product of speed and time. On the other hand, it is the area under the graph (see Fig. 25).
Rice. 25. Relationship between derivative and integral
If you calculate the integral of the speed, you get the value for the path. And speed is the derivative of the distance.
Therefore, the derivative and the integral are mutually inverse functions. There is strong evidence for this.
Rice. 26. Relationship between derivative and integral
But in order to analyze, understand what is at stake, and work with the operations of differentiation (calculating the derivative) and integration (calculating the integral), what has been said in this lesson and the materials from the main lessons will be enough.
When we need to find a house at st. Neva, and we went out in front of the house, then we go to the left or right of this house in order to understand how the numbering goes.
\(\DeclareMathOperator(\tg)(tg)\)\(\DeclareMathOperator(\ctg)(ctg)\)\(\DeclareMathOperator(\arctg)(arctg)\)\(\DeclareMathOperator(\arcctg)(arcctg) \)
ContentContent elements
Derivative, tangent, antiderivative, graphs of functions and derivatives.
Derivative Let the function \(f(x)\) be defined in some neighborhood of the point \(x_0\).
The derivative of the function \(f\) at the point \(x_0\) called the limit
\(f"(x_0)=\lim_(x\rightarrow x_0)\dfrac(f(x)-f(x_0))(x-x_0),\)
if this limit exists.
The derivative of a function at a point characterizes the rate of change of this function at a given point.
| Function | Derivative |
| \(const\) | \(0\) |
| \(x\) | \(1\) |
| \(x^n\) | \(n\cdot x^(n-1)\) |
| \(\dfrac(1)(x)\) | \(-\dfrac(1)(x^2)\) |
| \(\sqrt(x)\) | \(\dfrac(1)(2\sqrt(x))\) |
| \(e^x\) | \(e^x\) |
| \(a^x\) | \(a^x\cdot \ln(a)\) |
| \(\ln(x)\) | \(\dfrac(1)(x)\) |
| \(\log_a(x)\) | \(\dfrac(1)(x\ln(a))\) |
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(-\sin x\) |
| \(\tgx\) | \(\dfrac(1)(\cos^2 x)\) |
| \(\ctg x\) | \(-\dfrac(1)(\sin^2x)\) |
Differentiation rules\(f\) and \(g\) are functions depending on the variable \(x\); \(c\) is a number.
2) \((c\cdot f)"=c\cdot f"\)
3) \((f+g)"= f"+g"\)
4) \((f\cdot g)"=f"g+g"f\)
5) \(\left(\dfrac(f)(g)\right)"=\dfrac(f"g-g"f)(g^2)\)
6) \(\left(f\left(g(x)\right)\right)"=f"\left(g(x)\right)\cdot g"(x)\) - derivative of complex function
The geometric meaning of the derivative Equation of a straight line- non-parallel axis \(Oy\) can be written as \(y=kx+b\). The coefficient \(k\) in this equation is called slope of a straight line. It is equal to tangent tilt angle this straight line.
Straight angle- the angle between the positive direction of the \(Ox\) axis and the given line, counted in the direction of positive angles (that is, in the direction of least rotation from the \(Ox\) axis to the \(Oy\) axis).
The derivative of the function \(f(x)\) at the point \(x_0\) is equal to the slope of the tangent to the graph of the function at the given point: \(f"(x_0)=\tg\alpha.\)
If \(f"(x_0)=0\), then the tangent to the graph of the function \(f(x)\) at the point \(x_0\) is parallel to the axis \(Ox\).
Tangent equation
The equation of the tangent to the graph of the function \(f(x)\) at the point \(x_0\):
\(y=f(x_0)+f"(x_0)(x-x_0)\)
Function monotonicity If the derivative of a function is positive at all points in an interval, then the function is increasing on that interval.
If the derivative of a function is negative at all points in an interval, then the function is decreasing on that interval.
Minimum, maximum and inflection points positive on the negative at this point, then \(x_0\) is the maximum point of the function \(f\).
If the function \(f\) is continuous at the point \(x_0\), and the value of the derivative of this function \(f"\) changes from negative on the positive at this point, then \(x_0\) is the minimum point of the function \(f\).
The points at which the derivative \(f"\) is equal to zero or does not exist are called critical points functions \(f\).
Internal points of the function definition area \(f(x)\), where \(f"(x)=0\) can be minimum, maximum or inflection points.
The physical meaning of the derivative If a material point moves in a straight line and its coordinate changes depending on time according to the law \(x=x(t)\), then the speed of this point is equal to the time derivative of the coordinate:
The acceleration of a material point is equal to the derivative of the speed of this point with respect to time:
\(a(t)=v"(t).\)
This lesson is the first in a series of videos on integration. In it, we will analyze what the antiderivative of a function is, and also study the elementary methods for calculating these very antiderivatives.
In fact, there is nothing complicated here: in essence, everything comes down to the concept of a derivative, which you should already be familiar with. :)
I note right away that, since this is the very first lesson in our new topic, today there will be no complex calculations and formulas, but what we will study today will form the basis of much more complex calculations and structures when calculating complex integrals and areas.
In addition, when starting to study integration and integrals in particular, we implicitly assume that the student is already at least familiar with the concepts of the derivative and has at least elementary skills in calculating them. Without a clear understanding of this, there is absolutely nothing to do in integration.
However, here lies one of the most frequent and insidious problems. The fact is that, starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, stupid and offensive mistakes are made in exams and independent work.
Therefore, now I will not give a clear definition of antiderivative. And in return, I suggest you look at how it is considered on a simple concrete example.
What is primitive and how is it considered
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is considered elementary:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look closely at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can also write it this way, according to the definition of the derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of the antiderivative. But to write it correctly, you need to write the following:
Let's write the following expression in the same way:
If we generalize this rule, we can derive the following formula:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now we can formulate a clear definition.
An antiderivative of a function is a function whose derivative is equal to the original function.
Questions about the antiderivative function
It would seem that a fairly simple and understandable definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, well, this formula is correct. However, in this case, when $n=1$, we have problems: “zero” appears in the denominator, and it is impossible to divide by “zero”.
- The formula is only limited to powers. How to calculate the antiderivative, for example, sine, cosine and any other trigonometry, as well as constants.
- An existential question: is it always possible to find an antiderivative at all? If so, what about the antiderivative sum, difference, product, etc.?
I will answer the last question right away. Unfortunately, the antiderivative, unlike the derivative, is not always considered. There is no such universal formula, according to which from any initial construction we will obtain a function that will be equal to this similar construction. As for powers and constants, we'll talk about that now.
Solving problems with power functions
\[((x)^(-1))\to \frac(((x)^(-1+1)))(-1+1)=\frac(1)(0)\]
As you can see, this formula for $((x)^(-1))$ does not work. The question arises: what then works? Can't we count $((x)^(-1))$? Of course we can. Let's just start with this:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has been at least a little engaged in this topic will remember that this expression is equal to the derivative of the natural logarithm:
\[((\left(\ln x \right))^(\prime ))=\frac(1)(x)\]
Therefore, we can confidently write the following:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
This formula needs to be known, just like the derivative of a power function.
So what we know so far:
- For a power function — $((x)^(n))\to \frac(((x)^(n+1)))(n+1)$
- For a constant - $=const\to \cdot x$
- A special case of a power function - $\frac(1)(x)\to \ln x$
And if we begin to multiply and divide the simplest functions, how then to calculate the antiderivative of a product or a quotient. Unfortunately, analogies with the derivative of a product or a quotient do not work here. There is no standard formula. For some cases, there are tricky special formulas - we will get to know them in future video tutorials.
However, remember: general formula, there is no similar formula for calculating the derivative of a quotient and a product.
Solving real problems
Task #1
Let's each power functions count separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Task #2
As I have already said, primitive works and private "blank through" are not considered. However, here you can do the following:
We have broken the fraction into the sum of two fractions.
Let's calculate:
The good news is that once you know the formulas for computing antiderivatives, you are already able to calculate more complex structures. However, let's go ahead and expand our knowledge a little more. The fact is that many constructions and expressions that, at first glance, have nothing to do with $((x)^(n))$, can be represented as a degree with a rational exponent, namely:
\[\sqrt(x)=((x)^(\frac(1)(2)))\]
\[\sqrt[n](x)=((x)^(\frac(1)(n)))\]
\[\frac(1)(((x)^(n)))=((x)^(-n))\]
All these techniques can and should be combined. Power expressions can
- multiply (the powers are added);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving expressions with a degree with a rational exponent
Example #1
Let's count each root separately:
\[\sqrt(x)=((x)^(\frac(1)(2)))\to \frac(((x)^(\frac(1)(2)+1)))(\ frac(1)(2)+1)=\frac(((x)^(\frac(3)(2))))(\frac(3)(2))=\frac(2\cdot (( x)^(\frac(3)(2))))(3)\]
\[\sqrt(x)=((x)^(\frac(1)(4)))\to \frac(((x)^(\frac(1)(4))))(\frac( 1)(4)+1)=\frac(((x)^(\frac(5)(4))))(\frac(5)(4))=\frac(4\cdot ((x) ^(\frac(5)(4))))(5)\]
In total, our entire construction can be written as follows:
Example #2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore, we will get:
\[\frac(1)(((x)^(3)))=((x)^(-3))\to \frac(((x)^(-3+1)))(-3 +1)=\frac(((x)^(-2)))(-2)=-\frac(1)(2((x)^(2)))\]
In total, collecting everything in one expression, we can write:
Example #3
First, note that we have already calculated $\sqrt(x)$:
\[\sqrt(x)\to \frac(4((x)^(\frac(5)(4))))(5)\]
\[((x)^(\frac(3)(2)))\to \frac(((x)^(\frac(3)(2)+1)))(\frac(3)(2 )+1)=\frac(2\cdot ((x)^(\frac(5)(2))))(5)\]
Let's rewrite:
I hope I will not surprise anyone if I say that what we have just studied is only the simplest calculations of antiderivatives, the most elementary constructions. Let's now look at a little more complex examples, in which, in addition to tabular antiderivatives, you still need to remember the school curriculum, namely, the formulas for abbreviated multiplication.
Solving More Complex Examples
Task #1
Recall the formula for the square of the difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the antiderivative of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
We collect everything in a common design:
Task #2
In this case, we need to open the difference cube. Let's remember:
\[((\left(ab \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Given this fact, it can be written as follows:
Let's modify our function a bit:
We consider, as always, for each term separately:
\[((x)^(-3))\to \frac(((x)^(-2)))(-2)\]
\[((x)^(-2))\to \frac(((x)^(-1)))(-1)\]
\[((x)^(-1))\to \ln x\]
Let's write the resulting construction:
Task #3
On top we have the square of the sum, let's open it:
\[\frac(((\left(x+\sqrt(x) \right))^(2)))(x)=\frac(((x)^(2))+2x\cdot \sqrt(x )+((\left(\sqrt(x) \right))^(2)))(x)=\]
\[=\frac(((x)^(2)))(x)+\frac(2x\sqrt(x))(x)+\frac(x)(x)=x+2((x) ^(\frac(1)(2)))+1\]
\[((x)^(\frac(1)(2)))\to \frac(2\cdot ((x)^(\frac(3)(2))))(3)\]
Let's write the final solution:
And now attention! Very important thing, with which the lion's share of errors and misunderstandings is associated. The fact is that until now, counting antiderivatives with the help of derivatives, giving transformations, we did not think about what the derivative of a constant is equal to. But the derivative of a constant is equal to "zero". And this means that you can write the following options:
- $((x)^(2))\to \frac(((x)^(3)))(3)$
- $((x)^(2))\to \frac(((x)^(3)))(3)+1$
- $((x)^(2))\to \frac(((x)^(3)))(3)+C$
This is very important to understand: if the derivative of a function is always the same, then the same function has an infinite number of antiderivatives. We can simply add any constant numbers to our primitives and get new ones.
It is no coincidence that in the explanation of the tasks that we have just solved, it was written “Write down general form primitives." Those. it is already assumed in advance that there is not one, but a whole multitude of them. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks, we will correct what we have not completed.
Once again, we rewrite our constructions:
In such cases, one should add that $C$ is a constant — $C=const$.
In our second function, we get the following construction:
And the last one:
And now we really got what was required of us in the initial condition of the problem.
Solving problems on finding antiderivatives with a given point
Now, when we know about constants and about the peculiarities of writing antiderivatives, it quite logically arises next type problems when out of the set of all antiderivatives it is required to find one and only one that would pass through a given point. What is this task?
The fact is that all antiderivatives of a given function differ only in that they are shifted vertically by some number. And this means that no matter what point on the coordinate plane we take, one antiderivative will definitely pass, and, moreover, only one.
So, the tasks that we will now solve are formulated as follows: it is not easy to find the antiderivative, knowing the formula of the original function, but to choose exactly one of them that passes through a given point, the coordinates of which will be given in the condition of the problem.
Example #1
First, let's just calculate each term:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((x)^(3))\to \frac(((x)^(4)))(4)\]
Now we substitute these expressions into our construction:
This function must pass through the point $M\left(-1;4 \right)$. What does it mean that it passes through a point? This means that if instead of $x$ we put $-1$ everywhere, and instead of $F\left(x \right)$ - $-4$, then we should get the correct numerical equality. Let's do this:
We see that we have an equation for $C$, so let's try to solve it:
Let's write down the very solution we were looking for:
Example #2
First of all, it is necessary to open the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original structure will be written as follows:
Now let's find $C$: substitute the coordinates of the point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As a final chord to what we have just analyzed, I propose to consider two more challenging tasks containing trigonometry. In them, in the same way, it will be necessary to find antiderivatives for all functions, then choose from this set the only one that passes through the point $M$ on the coordinate plane.
Looking ahead, I would like to note that the technique that we will now use to find antiderivatives from trigonometric functions, in fact, is a universal technique for self-testing.
Task #1
Let's remember the following formula:
\[((\left(\text(tg)x \right))^(\prime ))=\frac(1)(((\cos )^(2))x)\]
Based on this, we can write:
Let's substitute the coordinates of point $M$ into our expression:
\[-1=\text(tg)\frac(\text( )\!\!\pi\!\!\text( ))(\text(4))+C\]
Let's rewrite the expression with this fact in mind:
Task #2
Here it will be a little more difficult. Now you will see why.
Let's remember this formula:
\[((\left(\text(ctg)x \right))^(\prime ))=-\frac(1)(((\sin )^(2))x)\]
To get rid of the "minus", you must do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Substitute the coordinates of the point $M$:
Let's write down the final construction:
That's all I wanted to tell you today. We studied the very term antiderivatives, how to count them from elementary functions, and also how to find an antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you a little to understand this difficult topic. In any case, it is on the antiderivatives that indefinite and indefinite integrals, so it is absolutely necessary to count them. That's all for me. See you soon!
The line y=3x+2 is tangent to the graph of the function y=-12x^2+bx-10. Find b , given that the abscissa of the touch point is less than zero.
Show SolutionSolution
Let x_0 be the abscissa of the point on the graph of the function y=-12x^2+bx-10 through which the tangent to this graph passes.
The value of the derivative at the point x_0 is equal to the slope of the tangent, i.e. y"(x_0)=-24x_0+b=3. On the other hand, the tangent point belongs to both the graph of the function and the tangent, i.e. -12x_0^2+bx_0-10= 3x_0 + 2. We get a system of equations \begin(cases) -24x_0+b=3,\\-12x_0^2+bx_0-10=3x_0+2. \end(cases)
Solving this system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the condition of the abscissa, the touch points are less than zero, therefore x_0=-1, then b=3+24x_0=-21.
Answer
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight line segments). Using the figure, compute F(9)-F(5), where F(x) is one of the antiderivatives of f(x).
Show SolutionSolution
According to the Newton-Leibniz formula, the difference F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid bounded by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. According to the graph, we determine that the specified curvilinear trapezoid is a trapezoid with bases equal to 4 and 3 and a height of 3.
Its area is equal to \frac(4+3)(2)\cdot 3=10.5.
Answer
Source: "Mathematics. Preparation for the exam-2017. Profile level". Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
The figure shows a graph of y \u003d f "(x) - the derivative of the function f (x), defined on the interval (-4; 10). Find the intervals of decreasing function f (x). In your answer, indicate the length of the largest of them.
Solution
As you know, the function f (x) decreases on those intervals, at each point of which the derivative f "(x) is less than zero. Considering that it is necessary to find the length of the largest of them, three such intervals are naturally distinguished from the figure: (-4; -2) ;(0;3);(5;9).
The length of the largest of them - (5; 9) is equal to 4.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
The figure shows a graph of y \u003d f "(x) - the derivative of the function f (x), defined on the interval (-8; 7). Find the number of maximum points of the function f (x) belonging to the interval [-6; -2].
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Solution
The graph shows that the derivative f "(x) of the function f (x) changes sign from plus to minus (there will be a maximum at such points) at exactly one point (between -5 and -4) from the interval [-6; -2 Therefore, there is exactly one maximum point on the interval [-6;-2].
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
The figure shows a graph of the function y=f(x) defined on the interval (-2; 8). Determine the number of points where the derivative of the function f(x) is equal to 0 .
Solution
If the derivative at a point is equal to zero, then the tangent to the graph of the function drawn at this point is parallel to the Ox axis. Therefore, we find such points at which the tangent to the function graph is parallel to the Ox axis. On this chart, such points are extremum points (maximum or minimum points). As you can see, there are 5 extremum points.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
The line y=-3x+4 is parallel to the tangent to the graph of the function y=-x^2+5x-7. Find the abscissa of the point of contact.
Show SolutionSolution
The slope of the line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is y"(x_0). But y"=-2x+5, so y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is -3.Parallel lines have the same slope coefficients.Therefore, we find such a value x_0 that =-2x_0 +5=-3.
We get: x_0 = 4.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
The figure shows a graph of the function y=f(x) and marked points -6, -1, 1, 4 on the x-axis. At which of these points is the value of the derivative the smallest? Please indicate this point in your answer.
File for lesson 29.
Derivative. Derivative application. Primitive.
The slope of the tangent to the graph of the function at the point with the abscissa x 0 equal to the derivative of the function at the point x 0. .
Those. the derivative of the function at the point x 0 is equal to the tangent of the slope of the tangent drawn to the graph of the function at the point (x 0; f (x 0)).
Exercise 1. The figure shows a graph of the function y \u003d f (x) and a tangent to this graph, drawn at a point with an abscissa x x 0 .
Answer: 0.25
Exercise 2. The figure shows a graph of the function y \u003d f (x) and a tangent to this graph, drawn at a point with an abscissa x 0 . Find the value of the derivative of the function f(x) at the point x 0 . Answer: 0.6
Exercise 3. The figure shows a graph of the function y \u003d f (x) and a tangent to this graph, drawn at a point with an abscissa x 0 . Find the value of the derivative of the function f(x) at the point x 0 . Answer: -0.25
Exercise 4. The figure shows a graph of the function y \u003d f (x) and a tangent to this graph, drawn at a point with an abscissa x 0 . Find the value of the derivative of the function f(x) at the point x 0 . Answer: -0.2.
mechanical sense derivative.
v ( t 0 ) = x' ( t 0 )
speed is the derivative of the coordinate on time. Likewise, acceleration is the derivative of speed with respect to time :
a = v' ( t ).
Exercise 5 . Material point moves rectilinearly according to the law x(t)=12 t 2 +4 t+27, where x is the distance from the reference point in meters, t is the time in seconds measured from the moment the movement started. Find its speed (in meters per second) at time t=2 s. Answer: 52
Task 6. The material point moves in a straight line according to the lawx (t) \u003d 16 t 3 + t 2 - 8 t + 180, where x- distance from the reference point in meters,t- time in seconds, measured since the start of the movement. At what point in time (in seconds) was her speed equal to 42 m/s? Answer: 1
Sufficient sign of increasing (decreasing) function
1. If f `(x ) at each point of the interval (, then the function increases by (.
2. If f `(x ) at each point of the interval (, then the function decreases by (.
Necessary condition extremum
If point x 0 is the extremum point of the function and at this point there is a derivative, then f `( x 0 )=0
Sufficient extremum condition
If f `( x 0 x 0 the value of the derivative changes sign from "+" to "-", then x 0 is the maximum point of the function.
If f `( x 0 ) = 0 and when passing through the point x 0 the value of the derivative changes sign from "-" to "+", then x 0 is the minimum point of the function.
Task 7. The figure shows a graph of the derivative of the function f(x), defined on the interval (−7; 10). Find the number of minimum points of a function f(x) on the segment [−3; eight].
Solution. The minimum points correspond to the points where the sign of the derivative changes from minus to plus. On the segment [−3; 8] the function has one minimum point x= 4. Hence, such a point is 1. Answer: 1.
Task 8. The figure shows a graph of a differentiable function y=f(x) and marked seven points on the x-axis: x1, x2, x3, x4, x5, x6, x7. At how many of these points is the derivative of the function f(x) negative? Answer: 3
Task 9. The figure shows a graph of a differentiable function y=f(x) defined on the interval (− 11 ; − 1). Find a point from the segment [− 7 ; − 2], in which the derivative of the function f(x) is equal to 0. Answer: -4
Task 10. The figure shows a graph of the function y=f′(x) - the derivative of the function f(x), defined on the interval (2 ; 13). Find the maximum point of the function f(x). Answer: 9
Task 11. The figure shows the graph y=f′(x) of the derivative of the function f(x) defined on the interval (− 3; 8). At what point of the segment [− 2; 3] function f(x) takes the smallest value? Answer: -2
Task 12. The figure shows a graph of y=f "(x) - the derivative of the function f(x) defined on the interval (− 2 ; 11). Find the abscissa of the point at which the tangent to the graph of the function y=f(x) is parallel to the abscissa axis or coincides with her Answer: 3
Task 13. The figure shows a graph of y=f "(x) - the derivative of the function f(x), defined on the interval (− 4 ; 6). Find the abscissa of the point at which the tangent to the graph of the function y=f(x) is parallel to the line y=3x or matches it.Answer: 5
Task 14. The figure shows a graph of y=f "(x) - the derivative of the function f(x) defined on the interval (− 4 ; 13). Find the number of points where the tangent to the graph of the function y=f(x) is parallel to the line y=− 2x−10 or equal to it Answer: 5
Task 15. The line y =5x -8 is tangent to the graph of the function 4x 2 -15x +c . Find c. O answer: 17.
antiderivative
antiderivative function F(x) for function f(x) is called a function derivative which is equal to the original function. F " ( x )= f ( x ).
Task 16. The figure shows a graph y=F (x) one of the antiderivatives of some function f(x) defined on the interval (1;13). Using the figure, determine the number of solutions to the equation f (x)=0 on the segment . Answer: 4
Task 17. The figure shows a graph y=F(x) of one of the antiderivatives of some function f(x) defined on the interval (− 7; 8). Using the figure, determine the number of solutions of the equation f(x)=0 on the interval . Answer:1
Task 18. The figure shows a graph y=F(x) of one of the antiderivatives of some function f(x) and eight points are marked on the x-axis: x1, x2, x3, x4, x5, x6, x7, x8. At how many of these points is the function f(x) negative? Answer: 3
Task 19. The figure shows a graph of some function y=f(x). The function F(x)=12x 3 −3x 2 +152x−92 is one of the antiderivatives of the function f(x). Find the area of the shaded figure. Answer: 592
Algorithm for finding extremum points
Find the scope of the function.
Find the derivative of a function f "( x)
Find the points where f "( x) = 0.
Mark on the real line the domain of the function and all zeros of the derivative.
Define sign derivativefor each interval. (For this we substitute the "convenient" value x from this interval to f "( x)).
Determine by the signs of the derivative the areas of increase and decrease of the function and draw conclusions about the presence or absence of an extremum and its nature ( max ormin ) at each of these points.
Task 20. Find the maximum point of the function y=(2x−1)cosx−2sinx+5, belonging to the interval(0 ; π/2). Answer: 0.5
Task 21.Find the maximum point of the functiony=. Answer: 6
Finding algorithm the largest and smallest value of the function on the segment
Task 22. Find the smallest value of the function y =x −6x +1 on the segment . Answer: -31
Task 23. Find the smallest value of the function y=8cosx+30x/π+19 on the interval [− 2π/3; 0]. Answer: -5
Additionally. one. Find the maximum point of the function y=(x−11) 2 ⋅e x − 7 .
2. Find the largest value of the function y=x 5 -5x 3 -20x on the segment [− 9 ; one]. Answer: 48