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
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 segment
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 is for the "middle group of kindergarten", then such nuances are probably overkill.
Separately, many thanks for the "I will solve the exam" to all employees - an excellent guide.
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 )
