Task.

The function y=f(x) is defined on the interval (-5; 6). The figure shows a graph of the function y=f(x). Find among the points x 1, x 2, ..., x 7 those points at which the derivative of the function f(x) is equal to zero. In response, write down the number of points found.

Solution:

The principle in solving this problem is this: there are three possible behavior of the function on this interval:

1) when the function increases (the derivative there is greater than zero)

2) when the function is decreasing (where the derivative is less than zero)

3) when the function does not increase or decrease (where the derivative is either zero or does not exist)

We are interested in the third option.

The derivative is equal to zero where the function is smooth and does not exist at the break points. Let's look at all these points.

x 1 - the function increases, which means the derivative f′(x) >0

x 2 - the function takes a minimum and is smooth, which means the derivative f ′(x) = 0

x 3 - the function takes a maximum, but at this point there is a break, which means derivative f ′(x) does not exist

x 4 - the function takes a maximum, but at this point there is a break, which means derivative f ′(x) does not exist

x 5 - derivative f ′(x) = 0

x 6 - the function increases, which means the derivative f′(x) >0

x 7 - the function takes a minimum and is smooth, which means derivative f ′(x) = 0

We see that f ′(x) = 0 at points x 2, x 5 and x 7, a total of 3 points.

Studying a function using its derivative. In this article we will analyze some tasks related to the study of the graph of a function. In such problems, a graph of the function y = f (x) is given and questions are raised related to determining the number of points at which the derivative of the function is positive (or negative), as well as others. They are classified as tasks on applying derivatives to the study of functions.

Solving such problems, and in general problems related to research, is possible only with a full understanding of the properties of the derivative for studying the graphs of functions and the derivative. Therefore, I strongly recommend that you study the relevant theory. You can study and also watch (but it contains a brief summary).

We will also consider problems where the derivative graph is given in future articles, don’t miss it! So, the tasks:

The figure shows a graph of the function y = f (x), defined on the interval (−6; 8). Define:

1. The number of integer points at which the derivative of the function is negative;

2. The number of points at which the tangent to the graph of the function is parallel to the straight line y = 2;

1. The derivative of a function is negative on intervals on which the function decreases, that is, on the intervals (−6; –3), (0; 4.2), (6.9; 8). They contain the integer points −5, −4, 1, 2, 3, 4, and 7. We get 7 points.

2. Direct y= 2 parallel to axisOhy= 2 only at extremum points (at points where the graph changes its behavior from increasing to decreasing or vice versa). There are four such points: –3; 0; 4.2; 6.9

Decide for yourself:

Determine the number of integer points at which the derivative of the function is positive.

The figure shows a graph of the function y = f (x), defined on the interval (−5; 5). Define:

2. The number of integer points at which the tangent to the graph of the function is parallel to the straight line y = 3;

3. The number of points at which the derivative is zero;

1. From the properties of the derivative of a function it is known that it is positive on the intervals on which the function increases, i.e. on the intervals (1.4; 2.5) and (4.4; 5). They contain only one integer point x = 2.

2. Direct y= 3 parallel to axisOh. The tangent will be parallel to the liney= 3 only at extremum points (at points where the graph changes its behavior from increasing to decreasing or vice versa).

There are four such points: –4.3; 1.4; 2.5; 4.4

3. The derivative is equal to zero at four points (at extremum points), we have already indicated them.

Decide for yourself:

Determine the number of integer points at which the derivative of the function f(x) is negative.

The figure shows a graph of the function y = f (x), defined on the interval (−2; 12). Find:

1. The number of integer points at which the derivative of the function is positive;

2. The number of integer points at which the derivative of the function is negative;

3. The number of integer points at which the tangent to the graph of the function is parallel to the straight line y = 2;

4. The number of points at which the derivative is zero.

1. From the properties of the derivative of a function it is known that it is positive on intervals on which the function increases, i.e. on the intervals (–2; 1), (2; 4), (7; 9) and (10; 11). They contain integer points: –1, 0, 3, 8. There are four of them in total.

2. The derivative of a function is negative on intervals on which the function decreases, that is, on the intervals (1; 2), (4; 7), (9; 10), (11; 12). They contain integer points 5 and 6. We get 2 points.

3. Direct y= 2 parallel to axisOh. The tangent will be parallel to the liney= 2 only at extremum points (at points where the graph changes its behavior from increasing to decreasing or vice versa). There are seven such points: 1; 2; 4; 7; 9; 10; eleven.

4. The derivative is equal to zero at seven points (at extremum points), we have already indicated them.

Moreover, an infinitesimal is an infinitesimal of a lower order than an infinitesimal.

Definition 3. If the ratio of two infinitesimals / tends to unity, i.e. lim / 1 , then they are infinitesimal and are called equivalent

tape infinitesimal and write.

Example 2.24. Let =x, = ln(1+ x), where x 0. Infinitesimal and equivalent, since

			ln(1x)			ln(1 x ) lim ln[(1 x )1/ x ].
				x 0 x

We present without derivation several equivalent infinitesimals, the use of which greatly simplifies the calculation of limits:

x sin x, x tan x, x arcsin x, x arctan x, x e x 1.

3. DIFFERENTIAL CALCULUS OF A FUNCTION OF ONE VARIABLE

3.1. Definition of derivative and its geometric meaning

The limit of the ratio of the increment of the function y to the increment of the argument x that caused this increment, at x 0, i.e.

			f(x0	x)f(x0)

called derivative of a function f(x) in terms of the independent variable x.

	Designated					The operation of finding a derivative is called
				dx.
		f(x),

vayut differentiation.

The angular coefficient of the tangent drawn to the curve y = f (x) at some point is equal to the value of the derivative of the function at this point. This is geometric meaning of derivative.

Theorem 2. The constant factor can be taken out of the sign of the production

noah, i.e. if y cf (x), where c = const, then
				cf(x) .
Theorem 3. Derivative of the sum of a finite number of differentiables
functions is equal to the sum of the derivatives of these functions,						those. if y u (x) v (x),
		u (x) v (x) .
Theorem 4. Derivative					works	two differentiable

functions is equal to the product of the derivative of the first function by the second plus the product of the derivative of the second function by the first, i.e. if y u v then

y u v v u .

Theorem 5. The derivative of the quotient of two differentiable functions is equal to a fraction in which the denominator is equal to the square of the denominator, and the numerator is the difference between the products of the derivative of the numerator and the denominator and the product

	water denominator to numerator, i.e. If

3.3. Derivative of a complex function

Let a complex function y=f (x) be given, i.e. such that it can be represented in the following form: y=F (u), u =φ (x) or y=F (φ (x)). In the expression y=F (u), the variable u is called an intermediate argument.

	Theorem. If u=φ (x) has a derivative u x (x) at some point x,
	the function F (u) has at	appropriate			u value	derivative
y u F (u), then the complex function y=F (φ (x)) at the specified point x also has
derivative, which is equal to					where instead of u	there must be
		y x Fu		(u) x (x),

the expression u=φ(x) is substituted.

3.4. Table of basic differentiation formulas

Let's combine all the basic formulas and rules of differentiation into one table.

	y const				y" 0.
	y xn,				y" nxn 1 .
	y x ,				y" 1.
	y sin x,				y " cos x .

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 the positive direction of the axis. 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 the positive direction of the axis. 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 USE problems. 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

Problem B9 gives a graph of a function or derivative from which you need to determine one of the following quantities:

1. The value of the derivative at some point x 0,
2. Maximum or minimum points (extremum points),
3. Intervals of increasing and decreasing functions (intervals of monotonicity).

The functions and derivatives presented in this problem are always continuous, making the solution much easier. Despite the fact that the task belongs to the section of mathematical analysis, even the weakest students can do it, since no deep theoretical knowledge is required here.

To find the value of the derivative, extremum points and monotonicity intervals, there are simple and universal algorithms - all of them will be discussed below.

Read the conditions of problem B9 carefully to avoid making stupid mistakes: sometimes you come across quite lengthy texts, but there are few important conditions that affect the course of the solution.

Calculation of the derivative value. Two point method

If the problem is given a graph of a function f(x), tangent to this graph at some point x 0, and it is required to find the value of the derivative at this point, the following algorithm is applied:

1. Find two “adequate” points on the tangent graph: their coordinates must be integer. Let's denote these points as A (x 1 ; y 1) and B (x 2 ; y 2). Write down the coordinates correctly - this is a key point in the solution, and any mistake here will lead to an incorrect answer.
2. Knowing the coordinates, it is easy to calculate the increment of the argument Δx = x 2 − x 1 and the increment of the function Δy = y 2 − y 1 .
3. Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the increment of the function by the increment of the argument - and this will be the answer.

Let us note once again: points A and B must be looked for precisely on the tangent, and not on the graph of the function f(x), as often happens. The tangent line will necessarily contain at least two such points - otherwise the problem will not be formulated correctly.

Consider points A (−3; 2) and B (−1; 6) and find the increments:
Δx = x 2 − x 1 = −1 − (−3) = 2; Δy = y 2 − y 1 = 6 − 2 = 4.

Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.

Task. 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 .

Consider points A (0; 3) and B (3; 0), find the increments:
Δx = x 2 − x 1 = 3 − 0 = 3; Δy = y 2 − y 1 = 0 − 3 = −3.

Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.

Task. 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 .

Consider points A (0; 2) and B (5; 2) and find the increments:
Δx = x 2 − x 1 = 5 − 0 = 5; Δy = y 2 − y 1 = 2 − 2 = 0.

It remains to find the value of the derivative: D = Δy/Δx = 0/5 = 0.

From the last example, we can formulate a rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of tangency is zero. In this case, you don’t even need to count anything - just look at the graph.

Calculation of maximum and minimum points

Sometimes, instead of a graph of a function, Problem B9 gives a graph of the derivative and requires finding the maximum or minimum point of the function. In this situation, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:

1. The point x 0 is called the maximum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≥ f(x).
2. The point x 0 is called the minimum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≤ f(x).

In order to find the maximum and minimum points from the derivative graph, just follow these steps:

1. Redraw the derivative graph, removing all unnecessary information. As practice shows, unnecessary data only interferes with the decision. Therefore, we mark the zeros of the derivative on the coordinate axis - and that’s it.
2. Find out the signs of the derivative on the intervals between zeros. If for some point x 0 it is known that f'(x 0) ≠ 0, then only two options are possible: f'(x 0) ≥ 0 or f'(x 0) ≤ 0. The sign of the derivative is easy to determine from the original drawing: if the derivative graph lies above the OX axis, then f'(x) ≥ 0. And vice versa, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
3. We check the zeros and signs of the derivative again. Where the sign changes from minus to plus is the minimum point. Conversely, if the sign of the derivative changes from plus to minus, this is the maximum point. Counting is always done from left to right.

This scheme only works for continuous functions - there are no others in Problem B9.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−5; 5]. Find the minimum point of the function f(x) on this segment.

Let's get rid of unnecessary information and leave only the boundaries [−5; 5] and zeros of the derivative x = −3 and x = 2.5. We also note the signs:

Obviously, at the point x = −3 the sign of the derivative changes from minus to plus. This is the minimum point.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7]. Find the maximum point of the function f(x) on this segment.

Let's redraw the graph, leaving only the boundaries [−3; 7] and zeros of the derivative x = −1.7 and x = 5. Let us note the signs of the derivative on the resulting graph. We have:

Obviously, at the point x = 5 the sign of the derivative changes from plus to minus - this is the maximum point.

Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−6; 4]. Find the number of maximum points of the function f(x) belonging to the segment [−4; 3].

From the conditions of the problem it follows that it is enough to consider only the part of the graph limited by the segment [−4; 3]. Therefore, we build a new graph on which we mark only the boundaries [−4; 3] and zeros of the derivative inside it. Namely, points x = −3.5 and x = 2. We get:

On this graph there is only one maximum point x = 2. It is at this point that the sign of the derivative changes from plus to minus.

A small note about points with non-integer coordinates. For example, in the last problem the point x = −3.5 was considered, but with the same success we can take x = −3.4. If the problem is compiled correctly, such changes should not affect the answer, since the points “without a fixed place of residence” do not directly participate in solving the problem. Of course, this trick won’t work with integer points.

Finding intervals of increasing and decreasing functions

In such a problem, like the maximum and minimum points, it is proposed to use the derivative graph to find areas in which the function itself increases or decreases. First, let's define what increasing and decreasing are:

1. A function f(x) is said to be increasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≤ f(x 2). In other words, the larger the argument value, the larger the function value.
2. A function f(x) is called decreasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≥ f(x 2). Those. A larger argument value corresponds to a smaller function value.

Let us formulate sufficient conditions for increasing and decreasing:

1. In order for a continuous function f(x) to increase on the segment , it is sufficient that its derivative inside the segment be positive, i.e. f’(x) ≥ 0.
2. In order for a continuous function f(x) to decrease on the segment , it is sufficient that its derivative inside the segment be negative, i.e. f’(x) ≤ 0.

Let us accept these statements without evidence. Thus, we obtain a scheme for finding intervals of increasing and decreasing, which is in many ways similar to the algorithm for calculating extremum points:

1. Remove all unnecessary information. In the original graph of the derivative, we are primarily interested in the zeros of the function, so we will leave only them.
2. Mark the signs of the derivative at the intervals between zeros. Where f’(x) ≥ 0, the function increases, and where f’(x) ≤ 0, it decreases. If the problem sets restrictions on the variable x, we additionally mark them on a new graph.
3. Now that we know the behavior of the function and the constraints, it remains to calculate the quantity required in the problem.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7.5]. Find the intervals of decrease of the function f(x). In your answer, indicate the sum of the integers included in these intervals.

As usual, let's redraw the graph and mark the boundaries [−3; 7.5], as well as zeros of the derivative x = −1.5 and x = 5.3. Then we note the signs of the derivative. We have:

Since the derivative is negative on the interval (− 1.5), this is the interval of decreasing function. It remains to sum all the integers that are inside this interval:
−1 + 0 + 1 + 2 + 3 + 4 + 5 = 14.

Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−10; 4]. Find the intervals of increase of the function f(x). In your answer, indicate the length of the largest of them.

Let's get rid of unnecessary information. Let us leave only the boundaries [−10; 4] and zeros of the derivative, of which there were four this time: x = −8, x = −6, x = −3 and x = 2. Let’s mark the signs of the derivative and get the following picture:

We are interested in the intervals of increasing function, i.e. such where f’(x) ≥ 0. There are two such intervals on the graph: (−8; −6) and (−3; 2). Let's calculate their lengths:
l 1 = − 6 − (−8) = 2;
l 2 = 2 − (−3) = 5.

Since we need to find the length of the largest of the intervals, we write down the value l 2 = 5 as an answer.