In problem B9, a graph of a function or derivative is given, from which it is required to determine one of the following quantities:
- The value of the derivative at some point x 0,
- High or low points (extremum points),
- Intervals of increasing and decreasing functions (intervals of monotonicity).
The functions and derivatives presented in this problem are always continuous, which greatly simplifies the solution. Despite the fact that the task belongs to the section of mathematical analysis, it is quite within the power of even the most weak students, 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.
Carefully read the condition of problem B9 in order not to make stupid mistakes: sometimes quite voluminous texts come across, but there are few important conditions that affect the course of the solution.
Calculation of the value of the derivative. Two point method
If the problem is given a graph of the 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:
- 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 the key point of the solution, and any mistake here leads to the wrong answer.
- 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 .
- Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the function increment by the argument increment - and this will be the answer.
Once again, we note: points A and B must be sought precisely on the tangent, and not on the graph of the function f(x), as is often the case. The tangent will necessarily contain at least two such points, otherwise the problem is formulated incorrectly.
Consider the points A (−3; 2) and B (−1; 6) and find the increments:
Δx \u003d x 2 - x 1 \u003d -1 - (-3) \u003d 2; Δy \u003d y 2 - y 1 \u003d 6 - 2 \u003d 4.
Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.
Task. The figure shows the graph of the function y \u003d f (x) and the 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 increments:
Δx \u003d x 2 - x 1 \u003d 3 - 0 \u003d 3; Δy \u003d y 2 - y 1 \u003d 0 - 3 \u003d -3.
Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.
Task. The figure shows the graph of the function y \u003d f (x) and the 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 increments:
Δx \u003d x 2 - x 1 \u003d 5 - 0 \u003d 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 the rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of contact is equal to zero. In this case, you don’t even need to calculate anything - just look at the graph.
Calculating High and Low Points
Sometimes instead of a graph of a function in problem B9, a derivative graph is given and it is required to find the maximum or minimum point of the function. In this scenario, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:
- The point x 0 is called the maximum point of the function f(x) if the following inequality holds in some neighborhood of this point: f(x 0) ≥ f(x).
- The point x 0 is called the minimum point of the function f(x) if the following inequality holds in some neighborhood of this point: f(x 0) ≤ f(x).
In order to find the maximum and minimum points on the graph of the derivative, it is enough to perform the following steps:
- Redraw the graph of the derivative, removing all unnecessary information. As practice shows, extra data only interfere with the solution. Therefore, we mark the zeros of the derivative on the coordinate axis - and that's it.
- 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. Conversely, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
- We again check the zeros and signs of the derivative. Where the sign changes from minus to plus, there is a 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 works only 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 segment [−5; 5]. Find the minimum point of the function f(x) on this segment.
Let's get rid of unnecessary information - we will leave only the borders [−5; 5] and the zeros of the derivative x = −3 and x = 2.5. 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 the zeros of the derivative x = −1.7 and x = 5. 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 segment [−6; 4]. Find the number of maximum points of the function f(x) that belong to the interval [−4; 3].
It follows from the conditions of the problem that it is sufficient to consider only the part of the graph bounded by the segment [−4; 3]. Therefore, we build a new graph, on which we mark only the boundaries [−4; 3] and the zeros of the derivative inside it. Namely, the points x = −3.5 and x = 2. We get:
On this graph, there is only one maximum point x = 2. It is in it 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 formulated correctly, such changes should not affect the answer, since the points "without a fixed place of residence" are not directly involved in solving the problem. Of course, with integer points such a trick will not work.
Finding intervals of increase and decrease of a function
In such a problem, like the points of maximum and minimum, it is proposed to find areas in which the function itself increases or decreases from the graph of the derivative. First, let's define what ascending and descending are:
- A function f(x) is called increasing on a segment if for any two points x 1 and x 2 from this segment the statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≤ f(x 2). In other words, the larger the value of the argument, the larger the value of the function.
- A function f(x) is called decreasing on a segment if for any two points x 1 and x 2 from this segment the statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≥ f(x 2). Those. a larger value of the argument corresponds to a smaller value of the function.
We formulate sufficient conditions for increasing and decreasing:
- 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.
- 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.
We accept these assertions without proof. Thus, we get a scheme for finding intervals of increase and decrease, which is in many ways similar to the algorithm for calculating extremum points:
- Remove all redundant information. On the original graph of the derivative, we are primarily interested in the zeros of the function, so we leave only them.
- 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 has restrictions on the variable x, we additionally mark them on the new chart.
- Now that we know the behavior of the function and the constraint, it remains to calculate the required value 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 decreasing function f(x). In your answer, write the sum of integers included in these intervals.
As usual, we redraw the graph and mark the boundaries [−3; 7.5], as well as the zeros of the derivative x = −1.5 and x = 5.3. Then we mark 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 segment [−10; 4]. Find the intervals of increasing function f(x). In your answer, write the length of the largest of them.
Let's get rid of redundant information. We leave only the boundaries [−10; 4] and zeros of the derivative, which this time turned out to be four: x = −8, x = −6, x = −3 and x = 2. Note the signs of the derivative and get the following picture:
We are interested in the intervals of increasing function, i.e. 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 it is required to find the length of the largest of the intervals, we write the value l 2 = 5 in response.
Sergei 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 attached 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 it be argued 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.
Support service
The value of the derivative at a point is not directly related to the increase of the function on the interval. Consider, for example, functions - they all increase 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 the points a and b, then it is increasing on the segment . Those. the point x=2 is included in the given interval.
Although, as a rule, increase and decrease is considered not on a segment, but on an interval.
But at the very 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), then we do not count the points of local extremum, but they enter into the intervals of increase (decrease).
Considering that the first part of the exam for " middle group kindergarten", then perhaps such nuances are too much.
Separately, thank you very much for "I will decide the exam" for all employees - an excellent benefit.
Sergei 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 the interval if the larger argument of the function corresponds to a larger/smaller value of the function. Such a definition does not use the concept of a 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 borders should be included. Let's say I agree with this. But look, please, at your solution to problem 7089. There, when specifying intervals of increase, the boundaries are not included. And that affects the response. Those. the solutions of tasks 6429 and 7089 contradict each other. Please clarify this situation.
Alexander Ivanov
Tasks 6429 and 7089 have completely different questions.
In one, there are intervals of increase, and in the other, there are intervals with a positive derivative.
There is no contradiction.
The extrema are included in the intervals of increase and decrease, but the points at which the derivative is equal to zero do not enter the intervals at 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 the point 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, the inclusion of the given point x=2 is a special process.
Usually, to avoid confusion, the inclusion of the ends of the intervals is said separately.
Alexander Ivanov
The function y=f(x) is called increasing on some interval if the larger value of the argument from this interval corresponds to the 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 that on the interval )
