Unified State Exam training assignments on derivatives. Application of derivatives in exam tasks



Geometric meaning of the derivative X Y 0 tangent α k – angular coefficient of the straight line (tangent) Geometric meaning of the derivative: if a tangent can be drawn to the graph of the function y = f(x) at the point with the abscissa, non-parallel to the y-axis, then it expresses the angular coefficient of the tangent, i.e. e. Since, then the equality of the straight line is true


X y If α 0. If α > 90°, then k 90°, then k 90°, then k 90°, then k 90°, then k title="х y If α 0. If α > 90°, then k


X y Task 1. The figure shows a graph of the function y = f(x) and a tangent to this graph drawn at the point with abscissa -1. Find the value of the derivative of the function f(x) at the point x =






Y x x0x The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0. Answer: -0.25








The figure shows a graph of the derivative of the function f(x), defined on the interval (-6;6). Find the intervals of increase of the function f(x). In your answer, indicate the sum of integer points included in these intervals. B =...



The derivative of a function is one of the difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.

This article explains in a simple and clear way what a derivative is and why it is needed.. We will not now strive for mathematical rigor in the presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of a function.

The figure shows graphs of three functions. Which one do you think is growing faster?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here's another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

The graph shows everything at once, isn’t it? Kostya’s income more than doubled in six months. And Grisha’s income also increased, but just a little. And Matvey’s income decreased to zero. The starting conditions are the same, but the rate of change of the function, that is derivative, - different. As for Matvey, his income derivative is generally negative.

Intuitively, we easily estimate the rate of change of a function. But how do we do this?

What we're really looking at is how steeply the graph of a function goes up (or down). In other words, how quickly does y change as x changes? Obviously, the same function at different points can have different derivative values ​​- that is, it can change faster or slower.

The derivative of a function is denoted .

We'll show you how to find it using a graph.

A graph of some function has been drawn. Let's take a point with an abscissa on it. Let us draw a tangent to the graph of the function at this point. We want to estimate how steeply the graph of a function goes up. A convenient value for this is tangent of the tangent angle.

The derivative of a function at a point is equal to the tangent of the tangent angle drawn to the graph of the function at this point.

Please note that as the angle of inclination of the tangent we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what a tangent to the graph of a function is. This is a straight line that has a single common point with the graph in this section, and as shown in our figure. It looks like a tangent to a circle.

Let's find it. We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite side to the adjacent side. From the triangle:

We found the derivative using a graph without even knowing the formula of the function. Such problems are often found in the Unified State Examination in mathematics under the number.

There is another important relationship. Recall that the straight line is given by the equation

The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. It expresses the geometric meaning of the derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the tangent angle.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point the function increases. A tangent to the graph drawn at point forms an acute angle; with positive axis direction. This means that the derivative at the point is positive.

At the point our function decreases. The tangent at this point forms an obtuse angle; with positive axis direction. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If a function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

What will happen at the maximum and minimum points? We see that at the points (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the tangent at these points is zero, and the derivative is also zero.

Point - maximum point. At this point, the increase in the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from “plus” to “minus”.

At the point - the minimum point - the derivative is also zero, but its sign changes from “minus” to “plus”.

Conclusion: using the derivative we can find out everything that interests us about the behavior of a function.

If the derivative is positive, then the function increases.

If the derivative is negative, then the function decreases.

At the maximum point, the derivative is zero and changes sign from “plus” to “minus”.

At the minimum point, the derivative is also zero and changes sign from “minus” to “plus”.

Let's write these conclusions in the form of a table:

increases maximum point decreases minimum point increases
+ 0 - 0 +

Let's make two small clarifications. You will need one of them when solving the problem. Another - in the first year, with a more serious study of functions and derivatives.

It is possible that the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This is the so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it remains positive as it was.

It also happens that at the point of maximum or minimum the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

How to find the derivative if the function is given not by a graph, but by a formula? In this case it applies
























































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Lesson type: repetition and generalization.

Lesson format: lesson-consultation.

Lesson objectives:

  • educational: repeat and generalize theoretical knowledge on the topics: “Geometric meaning of the derivative” and “Application of the derivative to the study of functions”; consider all types of B8 problems encountered on the Unified State Examination in mathematics; provide students with the opportunity to test their knowledge by solving problems independently; teach how to fill out the exam answer form;
  • developing: to promote the development of communication as a method of scientific knowledge, semantic memory and voluntary attention; the formation of such key competencies as comparison, juxtaposition, classification of objects, determination of adequate ways to solve an educational task based on given algorithms, the ability to act independently in situations of uncertainty, monitor and evaluate one’s activities, find and eliminate the causes of difficulties;
  • educational: develop students’ communicative competencies (communication culture, ability to work in groups); promote the development of the need for self-education.

Technologies: developmental education, ICT.

Teaching methods: verbal, visual, practical, problematic.

Forms of work: individual, frontal, group.

Educational and methodological support:

1. Algebra and the beginnings of mathematical analysis. 11th grade: textbook. For general education Institutions: basic and profile. levels / (Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin); edited by A. B. Zhizhchenko. – 4th ed. – M.: Education, 2011.

2. Unified State Exam: 3000 problems with answers in mathematics. All tasks of group B / A.L. Semenov, I.V. Yashchenko and others; edited by A.L. Semyonova, I.V. Yashchenko. – M.: Publishing house “Exam”, 2011.

3. Open task bank.

Equipment and materials for the lesson: projector, screen, PC for each student with a presentation installed on it, printout of a memo for all students (Annex 1) and score sheet ( Appendix 2) .

Preliminary preparation for the lesson: as homework, students are asked to repeat theoretical material from the textbook on the topics: “Geometric meaning of the derivative”, “Application of the derivative to the study of functions”; The class is divided into groups (4 people each), in each of which there are students of different levels.

Lesson explanation: This lesson is taught in 11th grade at the stage of repetition and preparation for the Unified State Exam. The lesson is aimed at repetition and generalization of theoretical material, at applying it to solving exam problems. Lesson duration - 1.5 hours .

This lesson is not attached to the textbook, so it can be taught while working on any teaching materials. This lesson can also be divided into two separate ones and taught as final lessons on the topics covered.

During the classes

I. Organizational moment.

II. Setting goals lesson.

III. Repetition on the topic “Geometric meaning of derivatives.”

Oral frontal work using a projector (slides No. 3-7)

Work in groups: solving problems with hints, answers, with teacher consultation (slides No. 8-17)

IV. Independent work 1.

Students work individually on a PC (slides No. 18-26), and enter their answers into the evaluation sheet. If necessary, you can consult a teacher, but in this case the student will lose 0.5 points. If the student completes the work earlier, he can choose to solve additional tasks from the collection, pp. 242, 306-324 (additional tasks are assessed separately).

V. Mutual verification.

Students exchange assessment sheets, check a friend’s work, and assign points (slide No. 27)

VI. Correction of knowledge.

VII. Repetition on the topic “Application of the derivative to the study of functions”

Oral frontal work using a projector (slides No. 28-30)

Work in groups: solving problems with hints, answers, with teacher consultation (slides No. 31-33)

VIII. Independent work 2.

Students work individually on a PC (slides No. 34-46), and enter their answers on the answer form. If necessary, you can consult a teacher, but in this case the student will lose 0.5 points. If the student completes the work earlier, he can choose to solve additional tasks from the collection, pp. 243-305 (additional tasks are assessed separately).

IX. Peer review.

Students exchange assessment sheets, check their friend’s work, and assign points (slide No. 47).

X. Correction of knowledge.

Students work again in their groups, discuss the solution, and correct mistakes.

XI. Summarizing.

Each student calculates their points and puts a grade on the score sheet.

Students submit to the teacher an assessment sheet and solutions to additional problems.

Each student receives a memo (slide No. 53-54).

XII. Reflection.

Students are asked to evaluate their knowledge by choosing one of the phrases:

  • I succeeded!!!
  • We need to solve a couple more examples.
  • Well, who came up with this math!

XIII. Homework.

For homework, students are asked to choose tasks from the collection, pp. 242-334, as well as from an open bank of tasks.

Let's imagine a straight road passing through a hilly area. That is, it goes up and down, but does not turn right or left. If the axis is directed horizontally along the road and vertically, then the road line will be very similar to the graph of some continuous function:

The axis is a certain level of zero altitude; in life we ​​use sea level as it.

As we move forward along such a road, we also move up or down. We can also say: when the argument changes (movement along the abscissa axis), the value of the function changes (movement along the ordinate axis). Now let's think about how to determine the “steepness” of our road? What kind of value could this be? It’s very simple: how much the height will change when moving forward a certain distance. Indeed, on different sections of the road, moving forward (along the x-axis) by one kilometer, we will rise or fall by a different number of meters relative to sea level (along the y-axis).

Let’s denote progress (read “delta x”).

The Greek letter (delta) is commonly used in mathematics as a prefix meaning "change". That is - this is a change in quantity, - a change; then what is it? That's right, a change in magnitude.

Important: an expression is a single whole, one variable. Never separate the “delta” from the “x” or any other letter! That is, for example, .

So, we have moved forward, horizontally, by. If we compare the line of the road with the graph of the function, then how do we denote the rise? Certainly, . That is, as we move forward, we rise higher.

The value is easy to calculate: if at the beginning we were at a height, and after moving we found ourselves at a height, then. If the end point is lower than the starting point, it will be negative - this means that we are not ascending, but descending.

Let's return to "steepness": this is a value that shows how much (steeply) the height increases when moving forward one unit of distance:

Let us assume that on some section of the road, when moving forward by a kilometer, the road rises up by a kilometer. Then the slope at this place is equal. And if the road, while moving forward by m, dropped by km? Then the slope is equal.

Now let's look at the top of a hill. If you take the beginning of the section half a kilometer before the summit, and the end half a kilometer after it, you can see that the height is almost the same.

That is, according to our logic, it turns out that the slope here is almost equal to zero, which is clearly not true. Just over a distance of kilometers a lot can change. It is necessary to consider smaller areas for a more adequate and accurate assessment of steepness. For example, if you measure the change in height as you move one meter, the result will be much more accurate. But even this accuracy may not be enough for us - after all, if there is a pole in the middle of the road, we can simply pass it. What distance should we choose then? Centimeter? Millimeter? Less is better!

In real life, measuring distances to the nearest millimeter is more than enough. But mathematicians always strive for perfection. Therefore, the concept was invented infinitesimal, that is, the absolute value is less than any number that we can name. For example, you say: one trillionth! How much less? And you divide this number by - and it will be even less. And so on. If we want to write that a quantity is infinitesimal, we write like this: (we read “x tends to zero”). It is very important to understand that this number is not zero! But very close to it. This means that you can divide by it.

The concept opposite to infinitesimal is infinitely large (). You've probably already come across it when you were working on inequalities: this number is modulo greater than any number you can think of. If you come up with the biggest number possible, just multiply it by two and you'll get an even bigger number. And infinity is even greater than what happens. In fact, the infinitely large and the infinitely small are the inverse of each other, that is, at, and vice versa: at.

Now let's get back to our road. The ideally calculated slope is the slope calculated for an infinitesimal segment of the path, that is:

I note that with an infinitesimal displacement, the change in height will also be infinitesimal. But let me remind you that infinitesimal does not mean equal to zero. If you divide infinitesimal numbers by each other, you can get a completely ordinary number, for example, . That is, one small value can be exactly times larger than another.

What is all this for? The road, the steepness... We’re not going on a car rally, but we’re teaching mathematics. And in mathematics everything is exactly the same, only called differently.

Concept of derivative

The derivative of a function is the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument.

Incrementally in mathematics they call change. The extent to which the argument () changes as it moves along the axis is called argument increment and is designated. How much the function (height) has changed when moving forward along the axis by a distance is called function increment and is designated.

So, the derivative of a function is the ratio to when. We denote the derivative with the same letter as the function, only with a prime on the top right: or simply. So, let's write the derivative formula using these notations:

As in the analogy with the road, here when the function increases, the derivative is positive, and when it decreases, it is negative.

Can the derivative be equal to zero? Certainly. For example, if we are driving on a flat horizontal road, the steepness is zero. And it’s true, the height doesn’t change at all. So it is with the derivative: the derivative of a constant function (constant) is equal to zero:

since the increment of such a function is equal to zero for any.

Let's remember the hilltop example. It turned out that it was possible to arrange the ends of the segment on opposite sides of the vertex in such a way that the height at the ends turns out to be the same, that is, the segment is parallel to the axis:

But large segments are a sign of inaccurate measurement. We will raise our segment up parallel to itself, then its length will decrease.

Eventually, when we are infinitely close to the top, the length of the segment will become infinitesimal. But at the same time, it remained parallel to the axis, that is, the difference in heights at its ends is equal to zero (it does not tend to, but is equal to). So the derivative

This can be understood this way: when we stand at the very top, a small shift to the left or right changes our height negligibly.

There is also a purely algebraic explanation: to the left of the vertex the function increases, and to the right it decreases. As we found out earlier, when a function increases, the derivative is positive, and when it decreases, it is negative. But it changes smoothly, without jumps (since the road does not change its slope sharply anywhere). Therefore, there must be between negative and positive values. It will be where the function neither increases nor decreases - at the vertex point.

The same is true for the trough (the area where the function on the left decreases and on the right increases):

A little more about increments.

So we change the argument to magnitude. We change from what value? What has it (the argument) become now? We can choose any point, and now we will dance from it.

Consider a point with a coordinate. The value of the function in it is equal. Then we do the same increment: we increase the coordinate by. What is the argument now? Very easy: . What is the value of the function now? Where the argument goes, so does the function: . What about function increment? Nothing new: this is still the amount by which the function has changed:

Practice finding increments:

  1. Find the increment of the function at a point when the increment of the argument is equal to.
  2. The same goes for the function at a point.

Solutions:

At different points with the same argument increment, the function increment will be different. This means that the derivative at each point is different (we discussed this at the very beginning - the steepness of the road is different at different points). Therefore, when we write a derivative, we must indicate at what point:

Power function.

A power function is a function where the argument is to some degree (logical, right?).

Moreover - to any extent: .

The simplest case is when the exponent is:

Let's find its derivative at a point. Let's recall the definition of a derivative:

So the argument changes from to. What is the increment of the function?

Increment is this. But a function at any point is equal to its argument. That's why:

The derivative is equal to:

The derivative of is equal to:

b) Now consider the quadratic function (): .

Now let's remember that. This means that the value of the increment can be neglected, since it is infinitesimal, and therefore insignificant against the background of the other term:

So, we came up with another rule:

c) We continue the logical series: .

This expression can be simplified in different ways: open the first bracket using the formula for abbreviated multiplication of the cube of the sum, or factorize the entire expression using the difference of cubes formula. Try to do it yourself using any of the suggested methods.

So, I got the following:

And again let's remember that. This means that we can neglect all terms containing:

We get: .

d) Similar rules can be obtained for large powers:

e) It turns out that this rule can be generalized for a power function with an arbitrary exponent, not even an integer:

(2)

The rule can be formulated in the words: “the degree is brought forward as a coefficient, and then reduced by .”

We will prove this rule later (almost at the very end). Now let's look at a few examples. Find the derivative of the functions:

  1. (in two ways: by formula and using the definition of derivative - by calculating the increment of the function);

Trigonometric functions.

Here we will use one fact from higher mathematics:

With expression.

You will learn the proof in the first year of institute (and to get there, you need to pass the Unified State Exam well). Now I’ll just show it graphically:

We see that when the function does not exist - the point on the graph is cut out. But the closer to the value, the closer the function is to. This is what “aims.”

Additionally, you can check this rule using a calculator. Yes, yes, don’t be shy, take a calculator, we’re not at the Unified State Exam yet.

So, let's try: ;

Don't forget to switch your calculator to Radians mode!

etc. We see that the smaller, the closer the value of the ratio to.

a) Consider the function. As usual, let's find its increment:

Let's turn the difference of sines into a product. To do this, we use the formula (remember the topic “”): .

Now the derivative:

Let's make a replacement: . Then for infinitesimal it is also infinitesimal: . The expression for takes the form:

And now we remember that with the expression. And also, what if an infinitesimal quantity can be neglected in the sum (that is, at).

So, we get the following rule: the derivative of the sine is equal to the cosine:

These are basic (“tabular”) derivatives. Here they are in one list:

Later we will add a few more to them, but these are the most important, since they are used most often.

Practice:

  1. Find the derivative of the function at a point;
  2. Find the derivative of the function.

Solutions:

Exponent and natural logarithm.

There is a function in mathematics whose derivative for any value is equal to the value of the function itself at the same time. It is called “exponent”, and is an exponential function

The base of this function - a constant - is an infinite decimal fraction, that is, an irrational number (such as). It is called the “Euler number”, which is why it is denoted by a letter.

So, the rule:

Very easy to remember.

Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:

In our case, the base is the number:

Such a logarithm (that is, a logarithm with a base) is called “natural”, and we use a special notation for it: we write instead.

What is it equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

  1. Find the derivative of the function.
  2. What is the derivative of the function?

Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Rules of differentiation

Rules of what? Again a new term, again?!...

Differentiation is the process of finding the derivative.

That's all. What else can you call this process in one word? Not derivative... Mathematicians call the differential the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let it be, or simpler.

Examples.

Find the derivatives of the functions:

  1. at a point;
  2. at a point;
  3. at a point;
  4. at the point.

Solutions:

Derivative of the product

Everything is similar here: let’s introduce a new function and find its increment:

Derivative:

Examples:

  1. Find the derivatives of the functions and;
  2. Find the derivative of the function at a point.

Solutions:

Derivative of an exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just exponents (have you forgotten what that is yet?).

So, where is some number.

We already know the derivative of the function, so let's try to reduce our function to a new base:

To do this, we will use a simple rule: . Then:

Well, it worked. Now try to find the derivative, and don't forget that this function is complex.

Happened?

Here, check yourself:

The formula turned out to be very similar to the derivative of an exponent: as it was, it remains the same, only a factor appeared, which is just a number, but not a variable.

Examples:
Find the derivatives of the functions:

Answers:

Derivative of a logarithmic function

It’s similar here: you already know the derivative of the natural logarithm:

Therefore, to find an arbitrary logarithm with a different base, for example:

We need to reduce this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:

Only now we will write instead:

The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:

Derivatives of exponential and logarithmic functions are almost never found in the Unified State Examination, but it will not be superfluous to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.

Imagine a small conveyor belt: two people are sitting and doing some actions with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second one ties it with a ribbon. The result is a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example of a complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.

We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.

In other words, a complex function is a function whose argument is another function: .

For the first example, .

Second example: (same thing). .

The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

We change variables and get a function.

Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

  1. We define the “internal” function and find its derivative.
  2. We define the “external” function and find its derivative.
  3. We multiply the results of the first and second points.

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

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If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

The straight 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 tangent point is less than zero.

Show solution

Solution

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 point x_0 is equal to the slope of the tangent, that is, y"(x_0)=-24x_0+b=3. On the other hand, the point of tangency belongs simultaneously to both the graph of the function and the tangent, that is, -12x_0^2+bx_0-10= 3x_0 + 2. We obtain 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 abscissa condition, the tangent points are less than zero, so 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 segments). Using the figure, calculate F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x).

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Solution

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 limited by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and height 3.

Its area is equal \frac(4+3)(2)\cdot 3=10.5.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of y=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.

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Solution

As is known, 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 4.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of y=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 (at such points there will be a maximum) at exactly one point (between -5 and -4) from the interval [-6; -2 ] Therefore, on the interval [-6; -2] there is exactly one maximum point.

Answer

Source: “Mathematics. Preparation for the Unified State 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 at which the derivative of the function f(x) is equal to 0.

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Solution

The equality of the derivative at a point to zero means that the tangent to the graph of the function drawn at this point is parallel to the Ox axis. Therefore, we find points at which the tangent to the graph of the function 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 Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The straight 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 tangent point.

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Solution

The angular coefficient of the straight line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same slope coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

We get: x_0 = 4.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of the function y=f(x) and points -6, -1, 1, 4 are marked on the abscissa. At which of these points is the derivative the smallest? Please indicate this point in your answer.