The point where the derivative of the function is zero. Derivative of a function. Geometric meaning of derivative. The concept of increasing, decreasing, maximum, minimum of a function

Sergey Nikiforov

If the derivative of a function is of constant sign on an interval, and the function itself is continuous on its boundaries, then the boundary points are added to both increasing and decreasing intervals, which fully corresponds to the definition of increasing and decreasing functions.

Farit Yamaev 26.10.2016 18:50

Hello. How (on what basis) can we say that at the point where the derivative is equal to zero, the function increases. Give reasons. Otherwise, it's just someone's whim. By what theorem? And also proof. Thank you.

Help Desk

The value of the derivative at a point is not directly related to the increase in the function over the interval. Consider, for example, functions - they are all increasing on the interval

Vladlen Pisarev 02.11.2016 22:21

If a function is increasing on the interval (a;b) and is defined and continuous at points a and b, then it is increasing on the interval . Those. point x=2 is included in this interval.

Although, as a rule, increase and decrease are considered not on a segment, but on an interval.

But at the point x=2, the function has a local minimum. And how to explain to children that when they are looking for points of increase (decrease), we do not count the points of local extremum, but enter into intervals of increase (decrease).

Considering that the first part of the Unified State Exam For " middle group kindergarten", then perhaps such nuances are too much.

Separately, many thanks to all the staff for “Solving the Unified State Exam” - an excellent guide.

Sergey Nikiforov

A simple explanation can be obtained if we start from the definition of an increasing/decreasing function. Let me remind you that it sounds like this: a function is called increasing/decreasing on an interval if a larger argument of the function corresponds to a larger/smaller value of the function. This definition does not use the concept of derivative in any way, so questions about the points where the derivative vanishes cannot arise.

Irina Ishmakova 20.11.2017 11:46

Good afternoon. Here in the comments I see beliefs that boundaries need to be included. Let's say I agree with this. But please look at your solution to problem 7089. There, when specifying increasing intervals, boundaries are not included. And this affects the answer. Those. the solutions to tasks 6429 and 7089 contradict each other. Please clarify this situation.

Alexander Ivanov

Tasks 6429 and 7089 have completely different questions.

One is about increasing intervals, and the other is about intervals with a positive derivative.

There is no contradiction.

The extrema are included in the intervals of increasing and decreasing, but the points in which the derivative is equal to zero are not included in the intervals in which the derivative is positive.

A Z 28.01.2019 19:09

Colleagues, there is a concept of increasing at a point

(see Fichtenholtz for example)

and your understanding of the increase at x=2 is contrary to the classical definition.

Increasing and decreasing is a process and I would like to adhere to this principle.

In any interval that contains the point x=2, the function is not increasing. Therefore inclusion given point x=2 is a special process.

Usually, to avoid confusion, inclusion of the ends of intervals is discussed separately.

Alexander Ivanov

A function y=f(x) is said to be increasing over a certain interval if a larger value of the argument from this interval corresponds to a larger value of the function.

At the point x=2 the function is differentiable, and on the interval (2; 6) the derivative is positive, which means on the interval . 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.

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.

The derivative of a function is one of difficult topics V 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 meaning derivative - 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 function graph 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 only one common point with a graph, 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 right triangle 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:

Let's make two small clarifications. You will need one of them when solving Unified State Exam 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