Математика. Материалы курса

First, let’s look at the graph and we see that all these properties are easy to read on the graph. Even parity, odd parity: we see that the graph is not symmetric with respect to the О_у axis - there is no even parity. It is not centrally symmetric with respect to the origin point – it is not odd. Ascending and descending is related to how the function behaves when moving from left to right. Bounded from below means that there is a horizontal line below which there are no points in the function graph. We’ll speak about the inverse function later.

Let’s turn to solving problems. So, the first task is to investigate the function for even parity, odd parity. Studying at school, you used this definition: f(-x)= f(x) – for an even function and f(-x)=- f(x) – for an odd function. Let’s note that in each of these equalities, we are talking about equality of functions. Two functions are equal if two conditions are met: first, their definition area is equal, and second, the function values coincide at all points in the definition area. These two points (see the video) exactly mean what was said earlier. We will use a strict definition of the concept of equality of functions.

Let’s figure it out. On the first point, we note that the domain of the function under consideration is the set of all real numbers, the set is symmetric with respect to the point 0. The first point is fulfilled for both even and odd functions. Point two in the definition says that at symmetric points relative to point 0, the even function has equal values, and the odd function has values with the opposite sign.

How can we prove that a function is neither even nor odd? Take a pair of symmetric points, e. g., х₁=1, х₂= -1. We find the value of the function at these points, f(x₁)= 0 и f(x₂)= 8. Note that the value at point -1 is not equal to the value at point 1, so the function is not even. On the other hand, although these two numbers are not opposite in sign, we conclude that the function is not odd. So, a strict justification of the fact was carried out.

Let’s move on to the next task. It is necessary to prove that the function decreases on a certain set. If we want to prove it by definition, we write down this definition and act in accordance with these instructions (see the video). We take arbitrary points х₁, х₂ from the specified set (-∞;2], that х₁ < х₂. We prove that f(x₁) > f(x₂).

The easiest way to do this is to consider the difference f(x₁) - f(x₂). In the course of reasoning and transformation, we get that f(x₁) - f(x₂) > 0.by definition, this means that the function strictly decreases over the considered interval (-∞;2].

Let’s turn to the next task. We need to prove that the function is bounded from below. By definition (see the video), we need to find a number a, in which all function values are greater than or equal to this number, i. e., it is necessary to evaluate f(x). To do this, we perform the transformations that are familiar to us as the allocation of the full square. After that, we note that we found a real number a=-1, which for all x from the domain of definition, the function value will be greater than or equal to a. We found that the function is limited. If we turn to the graph, the established fact means the following: below the line y=-1 passing through the vertex of the parabola, there are no points of the graph of this function.

The next problem is related to the concept of reversibility. As the restriction, we consider the restriction of the function on the interval (-∞; 2]. We proved that the function strictly decreases over the specified interval. If the function is strictly steady, it is invertible. Following the algorithm for finding the inverse function, we express x through y (see the video). And after re-assigning the variables, we have the inverse function g(x).

We should remember that the function f(x) is non-injective, irreversible, and the resulting function g(x) is strictly steady, injective, and invertible. This function is for injectively narrowing the function f(x) by the specified interval.