Oof complex functions. Range of permissible values ​​- ODZ

Function y=f(x) is such a dependence of the variable y on the variable x, when each valid value of the variable x corresponds to a single value of the variable y.

Function definition domain D(f) is the set of all acceptable values variable x.

Function Range E(f) is the set of all admissible values ​​of the variable y.

Graph of a function y=f(x) is a set of points on the plane whose coordinates satisfy a given functional dependence, that is, points of the form M (x; f(x)). The graph of a function is a certain line on a plane.

If b=0 , then the function will take the form y=kx and will be called direct proportionality.

D(f) : x \in R;\enspace E(f) : y \in R

Schedule linear function- straight.

The slope k of the straight line y=kx+b is calculated using the following formula:

k= tan \alpha, where \alpha is the angle of inclination of the straight line to the positive direction of the Ox axis.

1) The function increases monotonically for k > 0.

For example: y=x+1

2) The function decreases monotonically as k< 0 .

For example: y=-x+1

3) If k=0, then giving b arbitrary values, we obtain a family of straight lines parallel to the Ox axis.

For example: y=-1

Inverse proportionality

Inverse proportionality called a function of the form y=\frac (k)(x), where k is a non-zero real number

D(f) : x \in \left \( R/x \neq 0 \right \); \: E(f) : y \in \left \(R/y \neq 0 \right \).

Function graph y=\frac (k)(x) is a hyperbole.

1) If k > 0, then the graph of the function will be located in the first and third quarters of the coordinate plane.

For example: y=\frac(1)(x)

2) If k< 0 , то график функции будет располагаться во второй и четвертой координатной плоскости.

For example: y=-\frac(1)(x)

Power function

Power function is a function of the form y=x^n, where n is a non-zero real number

1) If n=2, then y=x^2. D(f) : x \in R; \: E(f) : y \in; main period of the function T=2 \pi

There are an infinite number of functions in mathematics. And each has its own character.) To work with a wide variety of functions you need single an approach. Otherwise, what kind of mathematics is this?!) And there is such an approach!

When working with any function, we present it with a standard set of questions. And the first, the most important question- This domain of definition of the function. Sometimes this area is called the set of valid argument values, the area where a function is specified, etc.

What is the domain of a function? How to find it? These questions often seem complex and incomprehensible... Although, in fact, everything is extremely simple. You can see for yourself by reading this page. Go?)

Well, what can I say... Just respect.) Yes! The natural domain of a function (which is discussed here) matches with ODZ of expressions included in the function. Accordingly, they are searched according to the same rules.

Now let’s look at a not entirely natural domain of definition.)

Additional restrictions on the scope of a function.

Here we will talk about the restrictions that are imposed by the task. Those. The task contains some additional conditions that the compiler came up with. Or the restrictions emerge from the very method of defining the function.

As for the restrictions in the task, everything is simple. Usually, there is no need to look for anything, everything is already said in the task. Let me remind you that the restrictions written by the author of the task do not cancel fundamental limitations of mathematics. You just need to remember to take into account the conditions of the task.

For example, this task:

Find the domain of a function:

on the set of positive numbers.

We found the natural domain of definition of this function above. This area:

D(f)=( -∞ ; -1) ∪ (-1; 2] ∪ ∪

In the verbal method of specifying a function, you need to carefully read the condition and find restrictions on the Xs there. Sometimes the eyes look for formulas, but the words whistle past the consciousness yes...) Example from the previous lesson:

The function is specified by the condition: each value of the natural argument x is associated with the sum of the digits that make up the value of x.

It should be noted here that we are talking only about the natural values ​​of X. Then D(f) instantly recorded:

D(f): x ∈ N

As you can see, the domain of a function is not such a complicated concept. Finding this region comes down to examining the function, writing a system of inequalities, and solving this system. Of course, there are all kinds of systems, simple and complex. But...

I'll open it little secret. Sometimes a function for which you need to find the domain of definition looks simply intimidating. I want to turn pale and cry.) But as soon as I write down the system of inequalities... And, suddenly, the system turns out to be elementary! Moreover, often, the more terrible the function, the simpler the system...

Moral: the eyes fear, the head decides!)

A function is a model. Let's define X as a set of values ​​of an independent variable // independent means any.

A function is a rule with the help of which, for each value of an independent variable from the set X, one can find a unique value of the dependent variable. // i.e. for every x there is one y.

From the definition it follows that there are two concepts - an independent variable (which we denote by x and it can take any value) and a dependent variable (which we denote by y or f (x) and it is calculated from the function when we substitute x).

FOR EXAMPLE y=5+x

1. Independent is x, which means we take any value, let x=3

2. Now let’s calculate y, which means y=5+x=5+3=8. (y depends on x, because whatever x we ​​substitute, we get the same y)

The variable y is said to functionally depend on the variable x and is denoted as follows: y = f (x).

FOR EXAMPLE.

1.y=1/x. (called hyperbole)

2. y=x^2. (called parabola)

3.y=3x+7. (called straight line)

4. y= √ x. (called parabola branch)

The independent variable (which we denote by x) is called the function argument.

Function Domain

The set of all values ​​that a function argument takes is called the function's domain and is denoted D(f) or D(y).

Consider D(y) for 1.,2.,3.,4.

1. D (y)= (∞; 0) and (0;+∞) //the entire set of real numbers except zero.

2. D (y)= (∞; +∞)//all number of real numbers

3. D (y)= (∞; +∞)//all number of real numbers

4. D (y) = .

All this shows how important it is to have ODZ.

Example 3

Find the ODZ expression x 3 + 2 x y − 4 .

Solution

Any number can be cubed. This expression does not have a fraction, so the values ​​of x and y can be any. That is, ODZ is any number.

Answer: x and y – any values.

Example 4

Find the ODZ of the expression 1 3 - x + 1 0.

Solution

It can be seen that there is one fraction where the denominator is zero. This means that for any value of x we ​​will get division by zero. This means that we can conclude that this expression is considered undefined, that is, it does not have any additional liability.

Answer: ∅ .

Example 5

Find the ODZ of the given expression x + 2 · y + 3 - 5 · x.

Solution

Availability square root indicates that this expression must be greater than or equal to zero. If it is negative, it has no meaning. This means that it is necessary to write an inequality of the form x + 2 · y + 3 ≥ 0. That is, this is the desired range of acceptable values.

Answer: set of x and y, where x + 2 y + 3 ≥ 0.

Example 6

Determine the ODZ expression of the form 1 x + 1 - 1 + log x + 8 (x 2 + 3) .

Solution

By condition, we have a fraction, so its denominator should not be equal to zero. We get that x + 1 - 1 ≠ 0. The radical expression always makes sense when greater than or equal to zero, that is, x + 1 ≥ 0. Since it has a logarithm, its expression must be strictly positive, that is, x 2 + 3 > 0. The base of the logarithm must also have a positive value and different from 1, then we add the conditions x + 8 > 0 and x + 8 ≠ 1. It follows that the desired ODZ will take the form:

x + 1 - 1 ≠ 0, x + 1 ≥ 0, x 2 + 3 > 0, x + 8 > 0, x + 8 ≠ 1

In other words, it is called a system of inequalities with one variable. The solution will lead to the following ODZ notation [ − 1, 0) ∪ (0, + ∞) .

Answer: [ − 1 , 0) ∪ (0 , + ∞)

Why is it important to consider DPD when driving change?

During identity transformations, it is important to find the ODZ. There are cases when the existence of ODZ does not occur. To understand whether a given expression has a solution, you need to compare the VA of the variables of the original expression and the VA of the resulting one.

Identity transformations:

• may not affect DL;
• may lead to the expansion or addition of DZ;
• can narrow the DZ.

Let's look at an example.

Example 7

If we have an expression of the form x 2 + x + 3 · x, then its ODZ is defined over the entire domain of definition. Even when bringing similar terms and simplifying the expression, the ODZ does not change.

Example 8

If we take the example of the expression x + 3 x − 3 x, then things are different. We have a fractional expression. And we know that division by zero is unacceptable. Then the ODZ has the form (− ∞, 0) ∪ (0, + ∞) . It can be seen that zero is not a solution, so we add it with a parenthesis.

Let's consider an example with the presence of a radical expression.

Example 9

If there is x - 1 · x - 3, then you should pay attention to the ODZ, since it must be written as the inequality (x − 1) · (x − 3) ≥ 0. It is possible to solve by the interval method, then we find that the ODZ will take the form (− ∞, 1 ] ∪ [ 3 , + ∞) . After transforming x - 1 · x - 3 and applying the property of roots, we have that the ODZ can be supplemented and everything can be written in the form of a system of inequalities of the form x - 1 ≥ 0, x - 3 ≥ 0. When solving it, we find that [ 3 , + ∞) . This means that the ODZ is completely written as follows: (− ∞, 1 ] ∪ [ 3 , + ∞) .

Transformations that narrow the DZ must be avoided.

Example 10

Let's consider an example of the expression x - 1 · x - 3, when x = - 1. When substituting, we get that - 1 - 1 · - 1 - 3 = 8 = 2 2 . If we transform this expression and bring it to the form x - 1 · x - 3, then when calculating we find that 2 - 1 · 2 - 3 the expression makes no sense, since the radical expression should not be negative.

One should adhere to identical transformations that the ODZ will not change.

If there are examples that expand on it, then it should be added to the DL.

Example 11

Let's look at the example of a fraction of the form x x 3 + x. If we cancel by x, then we get that 1 x 2 + 1. Then the ODZ expands and becomes (− ∞ 0) ∪ (0 , + ∞) . Moreover, when calculating, we already work with the second simplified fraction.

In the presence of logarithms, the situation is slightly different.

Example 12

If there is an expression of the form ln x + ln (x + 3), it is replaced by ln (x · (x + 3)), based on the property of the logarithm. From this we can see that ODZ from (0 , + ∞) to (− ∞ , − 3) ∪ (0 , + ∞) . Therefore, to determine the ODZ ln (x · (x + 3)) it is necessary to carry out calculations on the ODZ, that is, the (0, + ∞) set.

When solving, it is always necessary to pay attention to the structure and type of the expression given by the condition. If the definition area is found correctly, the result will be positive.

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