Function definition area with root. How to find the field definition area

In each function there are two variables - an independent variable and a dependent variable, the values \u200b\u200bof which depend on the values \u200b\u200bof an independent variable. For example, in the function y. = f.(x.) = 2x. + y. An independent variable is "x", and the dependent - "y" (in other words, "y" is a function from "x"). The permissible values \u200b\u200bof the independent variable "X" are called the field definition area, and the values \u200b\u200bof the dependent variable "y" are called the function of the function values.

Steps

Part 1

Determine the type of functions given to you. The field of function values \u200b\u200bare all values \u200b\u200bof "x" (deposited along the horizontal axis), which correspond to the values \u200b\u200bof "y". The function may be quadratic or containing fractions or roots. To find the function of defining a function, you must first determine the type of function.

1. Select the appropriate entry for the function definition area. The definition area is written in square and / or parentheses. Square bracket It is used in the case when the value enters the function of determining the function; If the value is not included in the definition area, a round bracket is used. If the function has several non-negative areas of definition, the "U" symbol is set between them.

   • For example, the [-2.10) u (10.2] definition area includes -2 and 2 values, but does not include a value of 10.

2. Build graph quadratic function. The schedule of such a function is a parabola, whose branches are directed or up, or down. Since Parabola increases or decreases throughout the axis x, the area of \u200b\u200bdetermining the quadratic function is all valid numbers. In other words, the area of \u200b\u200bdefinition of such a function is the set R (R denotes all valid numbers).

   • To better clarify the concept of the function, select any value "x", substitute it to the function and find the value "y". A pair of "x" and "y" values \u200b\u200bare a point with coordinates (x, y), which lies on the graph of the function.
   • Apply this point to the coordinate plane and do the described process with another value of "x".
   • Applying the coordinate plane several points, you will receive general view On the form of a graph of the function.

3. If the function contains a fraction, equate its denominator to zero. Remember that it is impossible to divide to zero. Therefore, equating the denominator to zero, you will find the values \u200b\u200bof "X" that are not included in the field definition area.

   • For example, find the field definition area F (x) \u003d (x + 1) / (x - 1).
   • Here is the denominator: (x - 1).
   • Equate the denominator to zero and find "x": x - 1 \u003d 0; x \u003d 1.
   • Write down the function definition area. The definition area does not include 1, that is, it includes all valid numbers with the exception of 1. Thus, the function of determining the function: (-∞, 1) U (1, ∞).
   • Recording (-∞, 1) U (1, ∞) is read like this: the set of all valid numbers except 1. The symbol of infinity ∞ means all the actual numbers. In our example, all valid numbers that are greater than 1 and less than 1 are included in the definition area.

4. If the function contains a square root, then the feeding expression should be greater than or equal to zero. Remember that the square root of negative numbers is not retrieved. Therefore, any value of "x", in which the feeding expression becomes negative, should be excluded from the function of determining the function.

   • For example, find the function of defining the function f (x) \u003d √ (x + 3).
   • Guardian expression: (x + 3).
   • The feeding expression should be greater than or equal to zero: (x + 3) ≥ 0.
   • Find "x": x ≥ -3.
   • The definition area of \u200b\u200bthis function includes a plurality of all valid numbers that are greater or equal to -3. Thus, the definition area: [-3, ∞).

Part 2

1. Make sure you have a quadratic function. The quadratic function has the form: AX 2 + BX + C: F (X) \u003d 2X 2 + 3X + 4. The graph of such a function is a parabola, the branches of which are directed or up, or down. There are various methods for finding the area of \u200b\u200bthe values \u200b\u200bof the quadratic function.

   • The easiest way to find the range of functions containing the root or fraction is to build a graph of such a function using a graphical calculator.

2. Find the coordinate "x" vertices of the graphics of the function. In the case of a quadratic function, find the coordinate "X" of the pearabol vertex. Remember that the quadratic function has the form: AX 2 + BX + C. To calculate the "X" coordinate, use the following equation: x \u003d -b / 2a. This equation is a derivative of the main square function and describes the tangent, the angular coefficient of which is zero (tangent to the top of the parabola parallel to the axis X).

   • For example, find the range of values \u200b\u200bof the 3x 2 + 6x -2 function.
   • Calculate the coordinate "x" of the vertex parabola: x \u003d -b / 2a \u003d -6 / (2 * 3) \u003d -1

3. Find the coordinate "U" the vertices of the function graphics. To do this, substitute the "x" coordinate function. The desired coordinate "y" is the limit value of the function of function values.

   • Calculate the coordinate "y": y \u003d 3x 2 + 6x - 2 \u003d 3 (-1) 2 + 6 (-1) -2 \u003d -5
   • The coordinates of the vertex of parabola of this function: (-1, -5).

4. Determine the direction of parabola, substituting into the function at least one value "x". Select any other "X" value and substitute it to the function to calculate the corresponding "y" value. If the found value "y" is more coordinates of the "U" parabola vertex, then Parabola is directed upwards. If the found value "y" is less than the coordinate "y" of the pearabol vertex, then the parabol is directed down.

   • Submold to the function x \u003d -2: y \u003d 3x 2 + 6x - 2 \u003d y \u003d 3 (-2) 2 + 6 (-2) - 2 \u003d 12 -12 -2 \u003d -2.
   • The coordinates of the point lying on Parabola: (-2, -2).
   • The found coordinates indicate that the parabola branches are directed upwards. Thus, the function of the function values \u200b\u200bincludes all values \u200b\u200bof "y", which are greater or equal to -5.
   • The range of values \u200b\u200bof this function: [-5, ∞)

5. The function of the values \u200b\u200bof the function is recorded similar to the field definition area. The square bracket is used in the case when the value enters the function of the function values; If the value is not included in the range of values, a round bracket is used. If the function has several non-measure areas of values, the "U" symbol is set between them.

   • For example, the value of [-2.10) U (10.2] includes -2 and 2 values, but does not include 10.
   • Round brackets are always used with infinity symbol ∞.

Very often, when task is performed, the problem arises, how to find the field definition area? Without it, it is not to do without the construction of graphs and with further study of the values \u200b\u200bof the function.

Concept of function definition area

The function of determining the function is the set of variable values \u200b\u200bof the function x, in which the function f (x) makes sense. And more precisely, the value of the variable X function will be said, in which F (x) may exist in reality. For example, it is proposed to consider the case when the function cannot exist at all. The first case we will look at when in expression. In the embodiment, when the fraction occurs, the denominator must not be zero, for a simple reason that such fractional expressions simply do not exist, since they eventually lead to zero value, and one of the golden arithmetic rules - you can not divide on zero.

With zero figured out, let's deal with the scrimony. What to find the field definition area, examples with the same fraction, and determine the value of the variable x, we need to learn fraction to zero, and, solving this equation, we will get the value of the variable x, which will be excluded from the solution area. The second example is when our function contains an even degree root. Here we have complete freedom of action, since when solving such a function, we obtain a positive response with any subcortex number, which will be further deleted from the function of determining the function. What can not be said about the root of an odd degree when we can only suit a positively guided number.

Examples of solutions

Another example when you need to find the area of \u200b\u200bdata definition of the function specified by logarithm. It's absolutely simple here, the region of determining the logarithm is all positive numbers. And to find the values \u200b\u200bof the variable, it is necessary to solve inequality for this logarithm. Where the porphmic expression will be negative. Need to take into account trigonometric functions, namely, arcxinus and arckosinus, which are determined at the interval [-1: 1]. To do this, you need to trace, so that the expression value indicated by these functions fell into a predetermined gap to us, and everything else boldly exclude from the values \u200b\u200bof the variable.

One example, how to find a function of the function definition, if the function contains, for example, a difficult fraction. Where, for example, the denominator will look like a root of Arksinus. In this case, it is necessary to highlight only those values \u200b\u200bof the variable in which the arxinus may exist, and already remove the value of the arxinus that is zero (as it comes to this example Announcer), the next step is to exclude all the negative values, for the simple reason that they do not suit the condition of the function of the feeding value. All the remaining values \u200b\u200bare the desired.

Suppose our function has the form y \u003d A / B, its definition area is all values \u200b\u200bexcept for zero. The value of the number A can be completely any. For example, find the area of \u200b\u200bdefinition data Y \u003d 3 / 2x-1, we need to find those values \u200b\u200bof x, in which the denominator of the fraction will not be riveted to us. To do this, equating the denominator to zero and find a solution, after which the answer is obtained equal to 0.5 (x: 2x - 1 \u003d 0; 2x \u003d 1; x \u003d ½; x \u003d 0.5) Following this, from the region Function definitions should be excluded to 0.5. In order to find the field of definition of the function, the decision must take into account that this expression should be either positive or equal to zero.

It is necessary to find the field definition area of \u200b\u200bexamples y \u003d √3x-9, based on the above condition, we transform our expression in the form of inequality 3x ≥ 9; x ≥ 3; 0, after the solution we come to the value that x is greater than or equal to 3, and we exclude all these values \u200b\u200bfrom the function of the function when determining the area of \u200b\u200bdetermining the function of the feeding expression with an odd indicator, it is necessary to take into account that in this case the value of x can be If the feeding expression is not fractional, and x not in the denominator. Example: y \u003d ³√2x-5, you can simply indicate that the variable x may be an absolutely any actual number. In how to find the field definition area in no case should forget that this number under logarithm must be positive.

Example: It is necessary to find the field of determining the data of the function y \u003d log2 (4x - 1). Considering the above condition, finding the value of this function should be calculated so, 4x - 1\u003e 0; From this it follows 4x\u003e 1; x\u003e 0.25. And the field of determining this function will be equal to all values \u200b\u200bgreater than 0.25.

Some sites offer to find the field of defining the function online and save time on finding solutions. Very convenient service, especially for students and students.

The function with a square root is defined only at the values \u200b\u200bof "X" when the guoked expression is nonnegative:. If the root is located in the denominator, the condition is obviously toughened :. Similar calculations are valid for any root of a positive degree: True, the root is already 4th degree in research functions I do not remember.

Example 5.

Decision: The past expression should be nonnegative:

Before continuing the decision, I remind the basic rules of work with inequalities, known from school.

I pay special attention! Inequalities are now considered with one variable - That is, there is only for us one axis dimension. Please do not confuse with inequalities of two variableswhere the entire coordinate plane is geometrically involved. However, there are pleasant coincidences! So, the following transformations are equivalent for inequality:

1) The components can be transferred from part to part with a sign change.

2) both parts of inequality can be multiplied by a positive number.

3) if both parts of inequality are multiplied by negative number, then you need to change sign of the inequality itself. For example, if it was "more", it will become "less"; If it was "less or equal," it will become "more either equal."

In the inequality, we will transfer the "Troika" to the right-hand side of the sign of the sign (Rule No. 1):

Multiply both parts of inequality on -1 (rule number 3):

Multiply both parts of inequality on (rule number 2):

Answer: domain:

The answer can also be recorded by an equivalent phrase: "The function is defined when".
Geometrically, the definition area is depicted by hatching corresponding intervals on the abscissa axis. In this case:

Once again I remind the geometric meaning of the field of definition - the graph of the function There is only on the shaded plot and is missing at.

In most cases, a purely analytical finding of the field of definition is suitable, but when the function is very troubled, the axis should be drawn and notes.

Example 6.

Find the field definition area

This is an example for an independent solution.

When under square root is a square twist or threefold, the situation is slightly complicated, and now we will analyze the solutions in detail:

Example 7.

Find the field definition area

Decision: The feeding expression should be strictly positive, that is, we need to solve inequality. In the first step, we are trying to decompose the square triple to multipliers:

Discriminant is positive, looking for roots:

Thus, parabola The abscissa axis is crossed at two points, which means that part of the parabola is located below the axis (inequality), and part of the parabola is above the axis (inequality we need).

Because the coefficient, the branches of parabola look up. From the foregoing it follows that inequality is performed at the intervals (the parabola branches go up to infinity), and the pearabol vertex is located at the interval below the abscissa axis, which corresponds to inequality:

! Note: if you are not fully understood by the explanation, please draw the second axis and the entire parabola! It is advisable to return to the article. Charts and properties of elementary functions and methods Hot Mathematics School Course Formulas.

Please note that the points themselves are inqualing (not included in the solution), since the inequality we have strict.

Answer: domain:

In general, many inequalities (including those considered) are solved by universal interval methodknown again from school program. But in cases of square two and three-tier, in my opinion, it is much more convenient and faster to analyze the location of the parabola relative to the axis. And the main method - the interval method we will analyze in detail in the article Zero function. Sign intervals.

Example 8.

Find the field definition area

This is an example for an independent solution. In the sample, the logic of the argument + the second way of solving and one more important transformation of inequality is commented in detail in detail, without knowing the student will chrome one leg ..., ... hmm ... at the expense of the leg, perhaps, got excited, rather - one finger. Thumb.

Can a function with square root be determined on the entire numeric line? Sure. Familiar all persons :. Or similar amount with exponential :. Indeed, for any meaning "X" and "ka":, therefore, it is also suppressed. For example, the function is defined on the entire numerical line. However, the function has a single point is still not included in the definition area, since they draw a denominator to zero. For the same reason for the function Points are excluded.

Some visitors of the site, the examples under consideration will seem elementary and primitive, but there is no chance - first, I try to "sharpen" the material for noobs, and secondly, I select realistic things under the coming tasks: full research Functions, finding areas of defining the function of two variablesand some others. Everything in mathematics clings to each other. Although the lovers of difficulties will also be left deprived, more solid tasks will meet here, and in the lesson
about interval method.