Instruction

1

The first step in finding the region

**of definition***of the expression*can make an exception for division by zero. If the expression contains a denominator which can vanish, you must find all the values at which it becomes zero, and exclude them.Example: 1/x. The denominator vanishes at x = 0. 0 will not be included in the scope**of the definition***of the expression*.(x-2)/((x^2)-3x+2). The denominator vanishes at x = 1 and x = 2. These values will not be included in the scope**of the definition***of the expression*.2

The expression may also contain a variety of irrationality. If

*the expression*consists of roots of even degree, radical*expression*must be non-negative.Examples: 2+v(x-4). Hence, x?4 - area**definition**of the given*expression*. x^(1/4) is the fourth root of x. Hence, x?0 - area**definition**of the given*expression*.3

In

*the expression*x involving logarithms, it is necessary to remember that the log base a is defined for α>0 except for a=1. The expression under the sign of logarithm must be greater than zero.4

If the expression contains functions of arcsine or arccosine, the value

*of the expression*under the sign of this function should be limited to -1 left and 1 right. Hence, you need to find the area**of the definition**of this*expression*.5

The expression may appear as division and square root. While region

**determine**all*of the expression*necessary to consider all factors that may lead to the restriction of this region. Deleting all unmatched values, you need to record area**definition**.**Region****definition**can accept any valid values in the absence of specific points.# Advice 2 : How to find the domain of the function solution

**Area**

**definition**

*of a function*is the set of values of the argument in which this function exists. Identify different ways of finding the area

**of definition**

*of the function*.

You will need

- - handle;
- paper

Instruction

1

Consider the scope

**of the definition**of some elementary functions. If the function has the form y = a/b a region**determination**are all values except zero. The number a is any number. For example, to find the region of**definition***of the function*y = 3/2x-1, it is necessary to find those values of x for which the denominator of the given fraction is not zero. To do this, find the values of x for which the denominator is equal to zero. For this Paranaita the denominator to zero and find the value, solving the resulting equation for x : 2x – 1 = 0; 2x = 1; x = ½; x = 0,5. It follows that the**definition***of the function*will be any number other than 0.5 in.2

To find the region of

**definition***of the function*radical expression with an even exponent, consider the fact that this expression must be greater than or equal to zero. Example: Find the area**of definition***of the function*y = √3x-9. Referring to the above condition, the expression takes the form of the inequality: 3x – 9 ≥ 0. Solve the following: 3x ≥ 9; x ≥ 3. So, the scope**of definition**of this*function*will be all values of x greater than or equal to 3, i.e., x ≥ 3.3

Finding the area

**of definition***of the function*radical expression with an odd exponent, you need to remember the rule that x can be any number if the radical expression is not a fraction. For example, to find the region of**definition***of the function*y = 3√2x-5 , it is sufficient to indicate that x is any real number.4

While the field

**definition**logarithmic*functions*, remember that the expression standing under the sign of the logarithm must be a positive value. For example, find the area**of definition***of the function*y = log2 (4x – 1). Given the above condition, find the area**of definition***of the function*in the following way: 4x – 1 > 0; hence 4 > 1; x > 0,25. Thus, the scope**of the definition***of the function*y = log2 (4x – 1) are all values of x > 0,25.# Advice 3 : How to find the area of allowable values

**Region**of admissible

**values**of the function should not be confused with the region

**of values**of the function. If the first is all x for which the equation or inequality can be solved, then the second is all the values of the function, that is. About the region of the allowed

**values**must always be remembered, as often the values x insidiously turn out together and therefore cannot be a solution of the equation.

You will need

- - an equation or inequality with a variable.

Instruction

1

Initially, as the tolerance

**values**, take the infinity. That is, imagine that the equation can be solved for all H. then, using some simple prohibitions of mathematics (it is impossible to divide by zero, the expression under the root of an even degree and the logarithm must be greater than zero), is excluded from DHS, an invalid variable value.2

If the variable x lies in the expression under the root of an even degree, put the condition that the expression under the root must be less than zero. Then solve this inequality, found the interval to exclude from the tolerance

**values**. Please note, it is not necessary to solve the entire equation – when searching for DHS, you decide only a small piece.3

Note the division sign. If the expression has a denominator containing a variable, Paranaita it to zero and solve the resulting equation. Exclude the value of a variable obtained from the region of the allowed

**values**.4

If the expression is a sign of the logarithm with variable base, be sure to put the following restriction: the base must always be greater than zero and not equal to one. If the variable is under the sign of logarithm, specify that all expression in brackets must be greater than unity. Decide received a small of the equation and eliminate invalid values from DHS.

5

If the equation or inequality multiple roots of even degree, operations division, or logarithms, find invalid

**values**separately for each expression. Then combine the solution by subtracting all the results obtained from the region of the allowed**values**.6

Even if you found DHS and obtained by solving the equation the roots satisfy him, it does not always mean that these values of x are solutions, so always check the solution by substitution. For example, try to solve the following equation: √ (2x-1)=-H. In the area of allowable

**values**here will include all numbers that satisfy 2-1≥0, i.e., x≥1/2. For the solution of the equation raise both parts in square, after simplifications, you get one root x=1. Please note, this root is included in the DPG, but after the substitution you will see that it is not a solution of the equation. Final answer – no roots.