Lecture 2. The study of the monotonicity of the function
Why do we need differential calculus? In order to explore functions, to know how they behave, to know as much detail as possible. That's what we need derivatives for.
We'll start with the study of the monotonicity of the function: increasing, decreasing, and constancy. These terms are related to the word monotonicity.
The first of these theorems is a necessary and sufficient condition for the constancy of a function on an interval. The function is constant on the interval if and only if the derivative of the function at any point on this interval is equal to zero.
On the one hand, it is obvious. If the function is constant, then its derivative is 0. We know this very well. It is more important to prove another point.
Let the derivative of the function f be identically zero on the interval (a, b) at any point. What should we do?
We need to prove that the function is constant, that at any two points on the interval (a, b) the values of the function coincide.
Let's take arbitrary two points x1 and x2 from the interval (a, b).
We assume that one of them is to the left of the other, let x1 < x2. On the segment [x1, x2], we can apply Lagrange's theorem to the function f. How can it be explained? There is a point x0 for which the Lagrange finite-increments formula is fulfilled.
Let’s analyze the situation.
The derivative is always equal to 0 on the interval (a, b), and hence at the point x0. Thus, the right side of the equation is zero, so the left side is zero too. In other words, at points x1 and x2, the function values are the same. The function is constant. The theorem is proved.
How can we apply this theorem, for example, to prove the identities that we sometimes use in solving problems of elementary mathematics and some other tasks?
Let's try to prove the first problem, and you can prove the second one yourself.
Let’s consider the function f(x)=arcsin x + arccos x. It is defined on the segment [-1, 1] and it is continuous. If you look at the table of derivatives (or rather remember the facts), you can easily check and make sure that the derivative of this function is 0, because the derivative of the summand functions differs only in signs.
By the theorem just proved, we conclude that this function is constant on the interval [-1, 1], but since it is also continuous over the entire segment, we can conclude that it is constant over the entire segment [-1, 1].
We have to find out this value of c. We take any point from the segment [-1, 1], for example, 0 and we get 1 – the value at the point 0 π/2. Since the function is constant it means that all its values are equal to π/2. So, the identity is proved. You will prove the second identity yourself.
Now let's consider how to study the function for monotonicity, for increasing and decreasing. This theorem is well known. If the derivative of the function f has a plus sign within the interval (a, b), the function f increases strictly on the interval (a, b). If the derivative has a minus sign, the function strictly decreases. Let’s prove the first point.
Let f’>0 on the interval (a, b). We have to prove that the function increases on this interval. Remember the definition. What does the expression "strictly increasing" mean? This means that whatever the points x1, x2 are, if x1 < x2, then the value of the function at these points is also associated with the inequation "less". Let's take some arbitrary points x1 and x2, x1 < x2.
We have to prove the inequation connecting f(x1) and f(x2). We apply Lagrange's theorem for [x1, x2]. There is a point x0 from this interval (x1, x2) at which the equality is fulfilled (see the video). Let's evaluate the right side. Both factors on the right side have a + sign, thus, the left side has a + sign too, that is, f(x1) < f(x2).It means that the function strictly increases.
That's what we need to prove.
You will prove point 2 in the same way yourself.
The value of the derivative can be 0 at some points, and at other points in the interval its value is less than 0 or greater than 0.
We can conclude that in these cases the function is still strictly monotone.
Look at the function, f (x)=x3, the value of the derivative is 3x2, it is greater than 0 at all points except 0, that is, we would conclude that the function strictly increases at the interval from minus infinity to 0 and at the second interval from 0 to plus infinity, and at point 0, the derivative is 0.
This observation leads us to the conclusion that the function strictly increases. Try to prove this fact.
Let's analyze the mistakes you sometimes make.
An easy example can illustrate the concept of monotonicity, and research on increasing and decreasing. We find the derivative.
Note that the function is defined everywhere except for the point 0.
Therefore, the derivative is less than 0 at the intervals from minus infinity to 0 and from 0 to plus infinity. There is strict decreasing on the two intervals. A mistake you should avoid. It often happens, but you cannot put a Union sign between these intervals.
The answer is written like this, not the Union sign between these intervals, but the conjunction "and" or a comma, since we list the intervals of monotonicity, the intervals of decreasing in this case.
The function is not decreasing at the union of sets.
Try to explain why this is not the case.
One more example and a small remark. The example. The function f(x)= x2.
It is clear that it strictly decreases on the interval from minus infinity to 0, and it strictly increases on the interval from 0 to plus infinity.
However, at point 0, the function is continuous, so we can add point 0 to the decreasing and increasing interval and write down the conclusion.
Thus, it is necessary to write both the interval and increasing and decreasing at the rays, which include point 0.
Let’s consider the next concept – the concept of extremum and the study of the function for the extremum. Let the function f be defined on a certain interval that contains the point x0 (this is important: this is neither the end of the segment nor the end of the interval). The point x0 is called a maximum point if the value at point x0 is greater than or equal to the value of the function at point x from the neighborhood of the point x0, some neighborhood. We take another inequation for the minimum.
If the strict inequation holds for all x not equal to x0, then we speak about the points of a strict maximum and minimum. The points of maximum and minimum are the points of extremum.
A necessary condition for the extremum. If we know that the point x0 is the extremum point of the function f, and f is differentiable at this point, then we conclude from this theorem that the derivative at the point x0 is 0.
The proof is quite simple.
Let's recall Fermat's theorem, one of the main theorems of differential calculus.
If x0 is the extremum point, then the function is defined on a certain interval containing the point x0 and it takes the largest and smallest values at this point. In general, the concept of an extremum is a local concept that characterizes the behavior of a function in a certain neighborhood.
This is exactly what the highest or lowest value means.
We apply Fermat's theorem and conclude that the derivative is 0. The theorem is proved.
In accordance with this theorem, we conclude that if the derivative is a number not equal to 0, then x0 cannot be an extremum point. This is a logical situation. Note that the derivative of the function f at a point is a limit, we calculate the limit by definition, then what are the situations?
This can be the number 0 or a number not equal to 0, or infinity, or it does not exist.
The first of these cases. The limit is a number that is not equal to 0. In accordance with the theorem just proved, we can conclude that x0 is not an extremum point.
What about other cases? If the derivative is 0 or infinite, or it does not exist. Is there an extremum or not? Let's try to figure it out.
Let's consider three cases: the derivative is 0, is equal to infinity, or it does not exist.
These three illustrations show us that x0 can be an extreme point.
The derivative is zero, it is a horizontal tangent; it is equal to infinity, it is a vertical tangent, and when the derivative does not exist, there is no tangent. In each of these cases, x0 can be the extremum point.
Let's look at more illustrations.
Can there be no extremum under the same three conditions?
Horizontal tangents, vertical tangents, and no tangents.
In each of these cases, there is no extremum of the function.
In the situation where the derivative is 0 or equal to infinity, or it does not exist, the point x0 is the point of a possible extremum.
There is a term for this.
This point is called a critical point.
We use critical points to distinguish the situations where the derivative is 0.
We also often call these points stationary.
This is a special case of a critical point.
Let's move on to the research. We have a critical point of the function f, and it is continuous and differentiable at that point. How can we determine whether there is an extremum there or not?
If the derivative f‘ changes its sign when passing through the point x0 to the opposite, then x0 is the extremum point.
If it changes its sign from plus to minus, it is a maximum point, if it changes its sign from minus to plus, it is the point of minimum. We won't prove it. I think you are able to prove this theorem yourself in accordance with the previous theorems (the necessary and sufficient condition for monotonicity).
Let’s consider the following concepts: the convexity of the curve and the inflection points. To do this, we need the knowledge of second-order derivatives. Let the function be differentiable at x0, that is, the graph is a smooth curve at x0.
The smoothness is not broken; the graph has a tangent. We say that the graph of a function (this curve) is convex up if near the tangent point the tangent lies above the graph of the function and it is convex down if the graph lies near the tangent point above the tangent.
The inflection point is characterized by the fact that if we build a tangent, it will split the graph into two parts: to the right of the point x0 – the tangents are on one side of the curve, and to the left of the point x0 – the tangents are on the other side of the curve. This point x0 is called an inflection point.
Thus, the inflection point separates segments with different directions of convexity.
How do we investigate a function for convexity? Consider the following. This theorem is very similar to the theorem on sufficient condition of monotonicity. In accordance with the latter: if the derivative has a plus sign, the function increases, if it has the minus sign the function decreases. In accordance with the former: if the second derivative has a + sign, then the function is convex down on the interval, if the sign of the second – order derivative is minus, then the function is convex up on the interval (a, b). The conclusions are slightly different, but the structure of the theorem is the same. How can we memorize it? There are two ways: plus denotes that the cup is filled or we have a smiling emoticon; minus or the sign of the second derivative denotes that the bowl has overturned, it's bad, the emoticon is not smiling.
The necessary inflection condition is formulated similarly to the necessary extremum condition.
However, the terms have changed a little. What do we know?
x0 is an inflection point, and the function f is twice differentiable at x0, that is, there is the second derivative at x0. The theorem allows us to conclude that the value of this second derivative at the point x0 is zero.
In this case (similarly to what we said about the first derivative), logical reasoning allows us to conclude: if the value of the second-order derivative is not 0, there is no inflection.
But if it is equal to 0 or infinity, or it does not exist, then in each of these cases, the point x0 may be an inflection point, or it may not be.
Therefore, we call the point x0 at which the second derivative is 0, equal to infinity, or does not exist the critical point of a function of the second kind.
Sometimes these points can also be called points of possible inflection or points that are suspicious of inflection.
How do you know which of them are inflection points and which are not? To do this we apply the theorem of sufficient condition of inflection. Let x0 be the critical point of the second kind of the function f, which is continuous at this point. Similarly to the extremum investigation: if the second derivative changes its sign to the opposite when passing through the point x0, then x0 is the inflection point.
Let's go back to the extremum study. We can apply the knowledge about the second derivative to the extremum of the function.
Let x0 be a stationary point, the point of a possible extremum.
The derivative is equal to 0. How can we find out if there is an extremum or not? If the second derivative is greater than 0 (we remember how the graph looks – the function is convex down at this point) it is a minimum; if the second derivative is less than 0, then x0 is a maximum point. It is clear for the function f(x)=x2 that the derivative is 0 at point 0 (a stationary point).
Why is this the minimum point? It is not necessary to build a numeric line or to determine the signs of the derivative. We need to consider the value of the second derivative. It is 2, it is greater than 0, so 0 is the minimum point.
However, it may turn out that the second derivative is also 0.
Then this theorem cannot be applied.
There is a theorem that allows you to find out an extremum using the value of higher-order derivatives. If it turns out that all derivatives of the function f up to n-1 order at the point x0 are equal to 0, and the nth derivative is not 0, then we come to the following conclusions.
If this value of n is an odd natural number, then there is no extremum; if it is even, then x0 is an extremum; if the value of the n-th order derivative at the point x0 is greater than 0 then it is the minimum point; if it is less than 0 it is the maximum point.
We have considered the function x2. If we consider the function f (x)=x4, it is a stationary point, the value of the derivative is 0 only at point 0.
We will find the value of the derivatives at point 0. What do we have?
All derivatives up to the third order at point 0 are equal to 0, but 4 is the number 24 and 4 is an even number, so there is an extremum.
The value of the derivative is greater than 0, so 0 is the minimum point