The sum of the modules is equal to the number. Extracurricular lesson - number module

Modulus of numbers this number itself is called if it is non-negative, or the same number with opposite sign, if it is negative.

For example, the modulus of the number 5 is 5, and the modulus of the number –5 is also 5.

That is, the modulus of a number is understood as an absolute value, absolute value this number without taking into account its sign.

Denoted as follows: |5|, | X|, |A| etc.

Rule:

Explanation:

|5| = 5
It reads like this: the modulus of the number 5 is 5.

|–5| = –(–5) = 5
It reads like this: the modulus of the number –5 is 5.

|0| = 0
It reads like this: the modulus of zero is zero.

Module properties:

Geometric meaning of the module.

The modulus of a number is the distance from zero to that number.

For example, let's take the number 5 again. The distance from 0 to 5 is the same as from 0 to –5 (Fig. 1). And when it is important for us to know only the length of the segment, then the sign has not only meaning, but also meaning. However, this is not entirely true: we measure distance only with positive numbers - or non-negative numbers. Let the division price of our scale be 1 cm. Then the length of the segment from zero to 5 is 5 cm, from zero to –5 is also 5 cm.

In practice, the distance is often measured not only from zero - the reference point can be any number (Fig. 2). But this does not change the essence. Notation of the form |a – b| expresses the distance between points A And b on the number line.

Example 1. Solve the equation | X – 1| = 3.

Solution .

The meaning of the equation is that the distance between points X and 1 is equal to 3 (Fig. 2). Therefore, from point 1 we count three divisions to the left and three divisions to the right - and we clearly see both values X:
X 1 = –2, X 2 = 4.

We can calculate it.

│X – 1 = 3
│X – 1 = –3

│X = 3 + 1
│X = –3 + 1

│X = 4
│ X = –2.

Answer : X 1 = –2; X 2 = 4.

Example 2. Find expression module:

Solution .

First, let's find out whether the expression is positive or negative. To do this, we transform the expression so that it consists of homogeneous numbers. Let's not look for the root of 5 - it's quite difficult. Let's do it simpler: let's raise 3 and 10 to the root. Then compare the magnitude of the numbers that make up the difference:

3 = √9. Therefore, 3√5 = √9 √5 = √45

10 = √100.

We see that the first number is less than the second. This means that the expression is negative, that is, its answer is less than zero:

3√5 – 10 < 0.

But according to the rule, the modulus of a negative number is the same number with the opposite sign. We have negative expression. Therefore, it is necessary to change its sign to the opposite one. The opposite expression for 3√5 – 10 is –(3√5 – 10). Let's open the brackets in it and get the answer:

–(3√5 – 10) = –3√5 + 10 = 10 – 3√5.

Answer .

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Numeric intervals

Neighborhood of a point Let x o be any real number (a point on the number line). A neighborhood of the point xo is any interval (a; b) containing the point x0. In particular, the interval (x o -ε, x o +ε), where ε >0, is called the ε-neighborhood of the point x o. The number xo is called the center.

3 QUESTION concept of a function A function is such a dependence of the variable y on the variable x, in which each value of the variable x corresponds to a single value of the variable y.

The variable x is called the independent variable or argument.

The variable y is called the dependent variable.

Methods for specifying a function

Tabular method. consists of specifying a table of individual argument values ​​and their corresponding function values. This method of defining a function is used when the domain of definition of the function is a discrete finite set.

With the tabular method of specifying a function, it is possible to approximately calculate the values ​​of the function that are not contained in the table, corresponding to intermediate values ​​of the argument. To do this, use the interpolation method.

The advantages of the tabular method of specifying a function are that it makes it possible to determine certain specific values ​​immediately, without additional measurements or calculations. However, in some cases, the table does not define the function completely, but only for some values ​​of the argument and does not provide a clear image of the nature of the change in the function depending on the change in the argument.

Graphic method. Function graph y = f(x) is the set of all points of the plane whose coordinates satisfy the given equation.

The graphical method of specifying a function does not always make it possible to accurately determine the numerical values ​​of the argument. However, it has a big advantage over other methods - visibility. In engineering and physics, a graphical method of specifying a function is often used, and a graph is the only way available for this.

In order for the graphical assignment of a function to be completely correct from a mathematical point of view, it is necessary to indicate the exact geometric design of the graph, which, most often, is specified by an equation. This leads to the following way of specifying a function.

Analytical method. To specify a function, you must specify a way in which, for each argument value, the corresponding function value can be found. The most common way to specify a function is using the formula y = f (x), where f (x) is some expression with the variable x. In this case, they say that the function is given by a formula or that the function is given analytically.

For an analytically defined function, the domain of definition of the function is sometimes not explicitly indicated. In this case, it is implied that the domain of definition of the function y = f (x) coincides with the domain of definition of the expression f (x), i.e., with the set of those values ​​of x for which the expression f (x) makes sense.

Natural domain of a function

Function Domain f- this is a lot X all argument values x, on which the function is specified.

To indicate the domain of definition of a function f a short notation of the form is used D(f).

explicit implicit parametric specification functions

If a function is given by the equation y=ƒ(x), resolved with respect to y, then the function is given in explicit form (explicit function).

Under implicit task functions understand the definition of a function in the form of an equation F(x;y)=0, not resolved with respect to y.

Anything obviously given function y=ƒ (x) can be written as implicitly given by the equation ƒ(x)-y=0, but not vice versa.

Equations with modules, solution methods. Part 1.

Before you begin directly studying techniques for solving such equations, it is important to understand the essence of the module, its geometric meaning. It is in understanding the definition of the module and its geometric meaning that the main methods for solving such equations are laid. The so-called method of intervals when opening modular brackets is so effective that using it it is possible to solve absolutely any equation or inequality with moduli. In this part we will study in detail two standard methods: interval method and method of replacing an equation with a set.

However, as we will see, these methods are always effective, but not always convenient and can lead to long and even not very convenient calculations, which naturally require more time to solve. Therefore, it is important to know those methods that significantly simplify the solution of certain equation structures. Squaring both sides of an equation, a method for introducing a new variable, graphic method, solving equations containing a modulus under the modulus sign. We will look at these methods in the next part.

Determination of the modulus of a number. Geometric meaning of the module.

First of all, let's get acquainted with geometric sense module:

Modulus of numbers a (|a|) call the distance on the number line from the origin (point 0) to the point A(a).

Based on this definition, let's look at some examples:

|7| - this is the distance from 0 to point 7, of course it is equal to 7. → | 7 |=7

|-5|- this distance from 0 to point -5 and it is equal to: 5. → |-5| = 5

We all understand that distance cannot be negative! Therefore |x| ≥ 0 always!

Let's solve the equation: |x |=4

This equation can be read like this: the distance from point 0 to point x is 4. Yeah, it turns out that from 0 we can move both to the left and to the right, which means moving to the left at a distance equal to 4 we will end up at the point: -4, and moving to the right we will end up at point: 4. Indeed, |-4 |=4 and |4 |=4.

Hence the answer is x=±4.

If you carefully study the previous equation, you will notice that: the distance to the right along the number line from 0 to the point is equal to the point itself, and the distance to the left from 0 to the number is equal to the opposite number! Understanding that the numbers to the right of 0 are positive, and the numbers to the left of 0 are negative, we formulate definition of the modulus of a number: modulus ( absolute value) numbers X(|x|) is the number itself X, if x ≥0, and the number is X, if x<0.

Here we need to find a set of points on the number line, the distance from 0 to which will be less than 3, let's imagine a number line, point 0 on it, go to the left and count one (-1), two (-2) and three (-3), stop. Next will be points that lie further than 3 or the distance to which from 0 is greater than 3, now we go to the right: one, two, three, stop again. Now we select all our points and get the interval x: (-3;3).

It is important that you see this clearly, if you still can’t, draw it on paper and look so that this illustration is completely clear to you, don’t be lazy and try to see the solutions to the following tasks in your mind:

|x |=11, x=? |x|=-5, x=?

|x |<8, х-? |х| <-6, х-?

|x |>2, x-? |x|> -3, x-?

|π-3|=? |-x²-10|=?

|√5-2|=? |2х-х²-3|=?

|x²+2|=? |x²+4|=0

|x²+3x+4|=? |-x²+9| ≤0

Did you notice the strange tasks in the second column? Indeed, the distance cannot be negative therefore: |x|=-5- has no solutions, of course it cannot be less than 0, therefore: |x|<-6 тоже не имеет решений, ну и естественно, что любое расстояние будет больше отрицательного числа, значит решением |x|>-3 are all numbers.

After you learn to quickly see pictures with solutions, read on.

Modulus of a rational number they call the distance from the origin to the point on the coordinate line corresponding to this number.

Since the distance (length of a segment) can only be expressed as a positive number or zero, we can say that the modulus of a number cannot be negative.

Module properties:

The modulus of a positive number is equal to the number itself.
|a| = a, if a > 0;

The modulus of a negative number is equal to the opposite number.
|-a| = a if a< 0;

Zero module equal to zero.
|0| = 0 if a = 0;

Opposite numbers have equal modules.
|-a| = |a|;

Examples of modules rational numbers:

4.Basic solution methods irrational equations and inequalities.

We call an equation or inequality irrational if it contains a variable under the radicals, that is, under the signs of the square, cube, etc. root. Irrational equations and inequalities have a certain specificity.

Let us recall that the range of permissible values ​​(abbreviated as VA) of an equation or inequality is the set of values ​​of a variable for which both sides given equation or inequalities make sense. In any task, you can do without searching for (and without mentioning) ODZ, so there is no particular need for this concept. But there is no harm in it either; Moreover, in certain situations, finding ODZ turns out to be very useful. Thus, in some irrational equations and inequalities it does not come down to any specific techniques - just a close look and taking into account the ODZ.

Equivalent transformations

We move on to consider standard types irrational equations and inequalities. Here, a preliminary search for DZ turns out to be, as a rule, an unnecessary step; These problems are most effectively solved with the help of appropriate equivalent transitions. Equations of the form √ A = √ B

Let's start with an example.

Suppose we need to solve the equation √ x = √ 2x + 1. Due to the monotonicity of the function √ x, the radical expressions must be equal: x = 2x+1, whence x = −1. However, substituting this x value into the equation gives negative numbers under radicals; therefore, x = −1 is not a root of this equation, and therefore it has no solutions. Now let's consider general situation. Let there be an equation √ A = √ B, where A and B are some expressions containing a variable. Then, firstly, the radical expressions must be equal: A = B. Secondly, both radical expressions must be non-negative; but by virtue of their equality, it is sufficient to require that one of them be non-negative. Thus, we have: √ A = √ B ⇔ (A = B, A > 0 or √ A = √ B ⇔ (A = B, B > 0. In this case, it is natural to require that the expression that is simpler is not negative.

5. Graphing a function, analytical expressions which the module contains:

The modulus of a number is the distance from the reference point to the point corresponding to this point.

Algorithm for plotting y=|f(x)|.

1.Build a graph y=f(x)

2. Leave the sections of the graph lying above the abscissa axis unchanged.

3. Areas lying below the x-axis are mirrored relative to this axis.

Algorithm for plotting y=f(|x|).

1. Let's build a graph y=f(x).

2. delete all points located to the left of the OY axis.

3. All points lying on the op-amp axis and to the right of it will be reflected symmetrically relative to the op-amp axis.

Algorithm for plotting |y|=|f(x)|

1.Build a graph y=f(x).

2. construct a graph y=|f(x)|.

3.Make it mirror image relative to the Ox axis.

6.Properties and schedule square function y=ax+bx+c

A function that can be specified by the formula y=ax2+bx+c, where a,b,c∈R and a≠0,

called a quadratic function.

The domain of definition of the function y=ax2+bx+c ( acceptable values arguments x) are all real numbers(R).

Schedule quadratic function is a parabola.

The abscissa of the vertex of a parabola (xo;yo) can be calculated using the formula:

To plot a quadratic function you need to:

1) calculate the coordinates of the vertex of the parabola: x0=−b/2a and y0, which is found by substituting the value x0 into function formula,

2) mark the vertex of the parabola on coordinate plane, draw the axis of symmetry of the parabola,

3) determine the direction of the branches of the parabola,

4) mark the point of intersection of the parabola with Oy axis,

5) create a table of values ​​by selecting required values argument x.

Having solved the quadratic equation ax2+bx+c=0, we obtain the points of intersection of the parabola with the Ox axis or the roots of the function (if the discriminant D>0)

if D<0, то точек пересечения параболы с осью Ox не существует,