Practical lesson 1. Studying the properties of functions using a derivative
The topic of the lesson is “Studying the properties of functions using a derivative.” Moreover, we will consider the first- and second-order derivative. Let’s look at such properties as monotonicity, extremum, convexity, and points of inflection.
The first task is to study the function for monotonicity and extremes. So, we see the function. The right-hand side is a fourth-order polynomial. Monotonicity and extremes are what we apply f’. Let’s start. We find the derivative. Since we need to find critical points, it is logical to find only stationary ones where the derivative is zero, we got the value of the derivative 4x3-4x. It is logical to factor it to determine the stationary points.
In addition, we will use the interval method for f’, so it is the factorization that will allow us to do this effectively. So, stationary points appeared. We mark them on the numeric line, then determine the sign f’ on each of the intervals. And we have this image (see the video). Look, where there is a minus, the arrow is down, where there is a plus, the arrow is up. In addition, these arrows showed us where the maximum is, where the minimum is. We have hatched all the points. This is the point where the derivative is zero, so these points are extremum points.
How do we write the correct answer? Let’s see. So, we write: f strictly decreases on the intervals since the function is continuous everywhere, we can add the ends of intervals to the intervals, and it strictly increases on intervals, we write it down without the union sign. Please note that you should not make such a mistake. Points -1 and 1 are the minimum points, and point 0 is the maximum point. The graph of this function looks like this (see the video). We did not set the task of building a graph, but we can see from the graph that all these properties are well read. And by the way, you can use this graph to understand why the function is not, for example, strictly decreasing on the union, strictly increasing on the union. Why is there this error?
The second task is to investigate the function for convexity and inflection. We have the same function. So, to solve this problem, we will need the same techniques, we will find the function f”. We have already found the first derivative, and we find the second one by differentiating it. We decompose it into multipliers and determine where the second-order derivative is zero. So, these are two points minus one divided by the root of three and one divided by the root of three. We call these point critical points of the second kind. What do we do? We mark the points on the numeric line and define the signs on each of the intervals for the function f”. Keep in mind, we used f’ to study it for extremum and monotonicity. We used the interval method for the function f’. To solve this problem, the interval method is used for the function f”. So, we write the function f” and its signs above the numeric line, and below the direct function f and its behavior. So, we draw the plus sign of the second derivative, the bowl is filled, the convexity is down, the minus sign, the bowl is overturned, convexity is up.
So, the conclusions are. Points minus one, divided by the root of three and with a plus sign, they are shaded. These are the points of the domain of definition, the function is continuous at these points, so these points turn out to be inflection points, the second derivative in them changes its sign. How do we write down the conclusion? Here, in contrast to monotonicity, we can write the union sign, we do not add the ends. At the points marked on the numeric line, the function has an inflection, and there are no convexities up or down in them. So, these are the features of the answer, pay attention to them. If we look at the graph, during some time, the direction of the convexity was down, then it changed up and then down again. The inflection points are not highlighted on the graph, we can only imagine where they are. Approximately, where the direction of the convexity has changed, they do not need to be marked in a special way on the graph of the function.
Let’s look at a text problem that also uses knowledge of the function study. The tin can has the shape of a cylinder of volume V, what should be its dimensions (height, radius of the base), so that it takes the least amount of tin to make it. In general, the problem shows that knowledge of differential computation is vital. How can the manufacturer solve this problem? Let’s help him.
So, the can has the shape of a cylinder, we denote the radius of the base r, and the height of the can h. So, we need to find out what r and h should be so that the surface area is the smallest. We make the formula of the square. The first term is the two areas of the base – the top and the bottom of the can, and 2 * π*r*h is the square of the rectangle, the sweep of the side surface. So, this value should be minimal. What’s wrong with this entry? Here, two variables r and h are involved at once, they change every time, and we would like to get a function of one variable. What do we use? Knowledge of the problem condition. We have a fixed volume of V. Let’s see how V is related to r and h. So, we see that h is expressed in terms of r. As a result, the function turns out to be a function of a single variable r. Let’s note that r since it is the radius of the base, this value is positive, there are no other restrictions here. So, the function is set in the range from 0 to plus infinity, and we need to find out at which r this function takes the smallest value. In other words, the problem is a completely abstract mathematical one.
So, the function S(r) is given, we examine it for an extremum. To do this, we find the derivative. We carefully transform this expression and see at which r the value of the derivative is zero. And so r0 is defined as the root of the third degree of V divided by 2π. A stationary point where the derivative is zero. How do we find out if the function S has an extremum at this point or not? And what is this extremum? What if there just be the highest value of the amount of tin that will be used to manufacture the can. What will we do? Let’s turn to the study of the extremum using the second derivative.
So, the second derivative is found, if we substitute r0, we get that the second derivative at the point r0 is not zero. This means that r0 is the extremum point. And the plus sign means that r0 is not only the extremum point, but also the minimum point. So, we can say what the size of the can should be. r0 is the value of the stationary point and h 0, we find the value of r0 and get the desired answer. So, we can advise the manufacturer to make cans with such values of r and h.