The values \u200b\u200bof the argument z. for which f.(z.) Turns to zero called. zero point. if a f.(a.) \u003d 0, then a - zero point.
ORD.Dot butnaz. zero ordern.
, if a
FCP can be represented as f.(z.) \u003d, where 
analytical function I. 
0.
In this case, in the decomposition of the function in a series of Taylor (43) the first n. coefficients are zero
=
=
Etc. Determine the order of zero for 
and (1 -COS z.) As z.
=
0
=
=
Zero 1 order
1 - COS. z.
=
=
zero 2 order
ORD. Dot z.
=
naz. infinitely remote pointand zero.functions f.(z.), if a f.(
) \u003d 0. Such a function is decomposed in a row on negative degrees z.
: f.(z.)
=
. If a
first n.
coefficients are zero, then we come to zero order n.
in an infinitely remote point: f.(z.)
= z.
-
n.
.
Isolated singular points are divided into: a) disposable singular points; b) poles ordern. ; at) significantly special points.
Dot but Naz. disposable special pointfunctions f.(z.) if z. 
a.
lim. f.(z.)
= with - Final .
Dot but Naz. polyce ordern.
(n.
1) Functions f.(z.) if the inverse function 
=
1/
f.(z.) has zero order n. At point but.This feature can always be represented as f.(z.)
=
where 
- analytical function and 
.
Dot but Naz. significantly special pointfunctions f.(z.) if z. 
a.
lim. f.(z.) does not exist.
Row Laurent.
Consider the case of the ring region of convergence r. < | z. 0 – a.| < R. with center at point but For function f.(z.). We introduce two new circles L. 1 (r.) I. L. 2 (R.) near the boundaries of the ring with a point z. 0 between them. We will make a split rings, on the edges of the cut by connecting the circle, moving to the one-connected area and in
cauchy Integral Formula (39) We obtain two integrals by variable Z
f.(z. 0)
=
+
,
(42)
where integration goes in opposite directions.
For integral in L. 1 Conditions | z. 0 – a. | > | z. – a. |, and for integral L. 2 Reverse Condition | z. 0 – a. | < | z. – a. |. Therefore, the factor 1 / ( z. – z. 0) Spread in a row (a) in the integral L. 2 and in a row (b) in the integral L. one . As a result, we get a decomposition f.(z.) in the annular region in row Laurent. on positive and negative degrees ( z. 0 – a.)
f.(z. 0)
=
A. n.
(z. 0 - A.) n.
(43)
where A. n.
=
=
;A. -N.
=
Planning for positive degrees (z. 0 - but) Naz. the right part A number of Laurent (a series of Taylor), and the decomposition of negative degrees is called. the main partrow of Laurent.
If inside the circle L. 1 There are no special points and the function analytic, then in (44) the first integral is zero on the Cauchy theorem and only the right part will remain in the decomposition of the function. Negative degrees in decomposition (45) appear only when analytically disappears within the inner circle and serve to describe a function near isolated singular points.
To build a row of Laurent (45) for f.(z.) You can calculate the decomposition coefficients for the general formula or use the decomposition of the elementary functions included in f.(z.).
The number of terms ( n.) The main part of the Laurent series depends on the type of high point: removing a special point
(n.
=
0)
; significantly special point
(n. 
);
polen. Order(n.
-
final).
and for f.(z.)
=
dot z.
= 0 removing a special pointbecause The main part is not. f.(z.)
=
(z.
-
) = 1 -
b) for f.(z.)
=
dot z.
= 0 -
Pole 1st order
f.(z.)
=
(z.
-
) =
-
c) for f.(z.) = e. 1 / z. dot z. = 0 - significantly special point
f.(z.)
=
e. 1 /
z. =
If a f.(z.) analytic in the area D. with the exception of m. Isolated singular points 
and | z. 1 |
< |z. 2 |
< . . . < |z. m. | then when decomposing a function in degrees z. The whole plane is divided into m. + 1 Ring | z. i. |
< | z.
| < | z. i. + 1 | And the Laurent series has a different view for each ring. When decomposition in degrees ( z.
–
z. i.
) Region of the convergence of the Laurent series is a circle | z.
–
z. i.
| < r.where r.
- Distance to the nearest high point.
Etc. Spread function f.(z.)
=
in a row of Laurent in degrees z. and ( z.
-
1).
Decision. Imagine the function in the form f.(z.)
= - z. 2
. We use the formula for the amount of geometric progression 
. In a circle | z |< 1 ряд сходится и f.(z.)
= - z. 2
(1 + z.
+ z. 2
+ z. 3
+ z. 4
+ . . .) = - z. 2
- z. 3
- z. four - . . . . Decomposition contains only right part. Let's go to the outer area of \u200b\u200bthe circle | z | \u003e 1. The function will be presenting as 
where 1 / | z.|
< 1, и получим разложение f.(z.)
= z.
=z.
+ 1 +
Because , decomposition of the function in degrees ( z.
-
1) has the form f.(z.)
= (z.
-
1) -1
+ 2 + (z.
-
1) for all 
1.
Etc. Decompose in a row of Laurent function f.(z.)
=
:
a) in degrees z. in a circle | z.|
< 1; b)
по степеням z.
ring 1.<
|z.|
< 3 ; c)
по степеням (z.
–
2). The presentation. Spider on the simplest fraction
=
=
+
=
.
From conditions z.
=1
A.
= -1/2 , z.
=3
B.
= ½.
but) f.(z.)
=
½ [ 
]
= ½ [ 
-(1/3)
], with | z.|<
1.
b) f.(z.)
= - ½ [ 
+
]
= -
(
), at 1< |z.|
< 3.
with) f.(z.)
=
½ [ 
]= -
½
[ 
]
=
=
- ½
= -
, with | 2 - z.|
< 1
This is a circle of radius 1 with the center at the point z. = 2 .
In some cases, power series can be reduced to the set of geometric progressions and after that it is easy to determine the region of their convergence.
Etc. Explore the convergence of the row
.
. . +
+
+
+
1
+ ()
+ ()
2
+ ()
3
+ . . .
Decision. This is the sum of two geometric progressions with q. 1
=
, q. 2 \u003d (). From the conditions of their convergence follows 
< 1 ,
< 1 или |z.|
> 1 , |z.|
< 2 , т.е. область сходимости ряда кольцо
1 < |z.|
< 2 .
Function - This is one of the most important mathematical concepts. Function - variable dependence w. from the variable x.if each value h. corresponds to the only value w.. Variable h. Called an independent variable or argument. Variable w. Called the dependent variable. All values \u200b\u200bof an independent variable (variable x.) Form the function of determining the function. All values \u200b\u200bthat adopted dependent variable (variable y.) Form the area of \u200b\u200bthe function values.
Graph graph Call the set of all points of the coordinate plane, the abscissions of which are equal to the values \u200b\u200bof the argument, and the ordinates - the corresponding values \u200b\u200bof the function, even on the abscissa axis, the values \u200b\u200bof the variable are postponed x., and the values \u200b\u200bof the variable are postponed along the ordinate axis y.. To build a graph function, you need to know the properties of the function. The main properties of the function will be discussed below!
To build a graph, we advise you to use our program - building graphs of functions online. If you have questions about studying the material on this page, you can always ask them on our forum. Also on the forum you will be helped to solve problems in mathematics, chemistry, geometry, probability theory and many other subjects!
The main properties of functions.
1) Function definition area and function values.
The function of determining the function is the set of all valid valid values \u200b\u200bof the argument. x. (variable x.), in which the function y \u003d f (x) Defined.
The range of function values \u200b\u200bis a set of all valid values. y.that takes the function.
In elementary mathematics, functions are studied only on a plurality of valid numbers.
2) zeros function.
Zero function is the value of the argument at which the function value is zero.
3) intervals of the symbol function.
The intervals of the functions of the function are such a plurality of the argument values \u200b\u200bon which the values \u200b\u200bof the function are only positive or only negative.
4) monotony.
The increasing function (in some interval) is a function that has a greater value of the argument from this gap corresponds to a greater value of the function.
A decreasing function (in some interval) is a function that has a greater value of the argument from this gap corresponds to a smaller value of the function.
5) parity (oddness) functions.
An even function is a function that the area of \u200b\u200bdetermining is symmetrical relative to the start of the coordinates and for any h. Equality is performed from the definition area f (-x) \u003d f (x). The chart of an even function is symmetrical about the axis of the ordinate.
An odd function is a function that has a field of determining symmetrical relative to the start of coordinates and for any h. Equality is right from the definition area f (-x) \u003d - f (x). The schedule of an odd function is symmetrical on the start of the coordinates.
6) limited and unlimited functions.
The function is called limited if there is a positive number M, which | F (x) | ≤ M for all x values. If there is no such number, the function is unlimited.
7) Periodic function.
The function f (x) is periodic, if there is such an excellent number T, which for any x f (x + t) \u003d f (x). Such a smallest number is called a function period. All trigonometric functions are periodic. (Trigonometric formulas).
After studying these properties of the function, you can easily explore the function and the properties of the function can build a function schedule. Also look at the material about the truth table, multiplication table, Mendeleev Table, Table of derivatives and integral table.
Zero function
What is zero functions? How to determine zeros of functions analytically and on schedule?
Zero function - These are the values \u200b\u200bof the argument in which the function is zero.
To find the zeros of the function given by the formula y \u003d f (x), it is necessary to solve the equation f (x) \u003d 0.
If the equation does not have roots, there is no zeros from the function.
1) Find the zeros of the linear function y \u003d 3x + 15.
To find the zeros of the functions, solve equation 3x + 15 \u003d 0.
Thus, zero functions y \u003d 3x + 15 - x \u003d -5.
2) Find the zeros of the quadratic function f (x) \u003d x²-7x + 12.
To find the zeros of the functions, the square equation
Its roots x1 \u003d 3 and x2 \u003d 4 are zeros of this function.
3) find zeros of functions
The fraction makes sense if the denominator is different from zero. Consequently, x²-1 ≠ 0, x² ≠ 1, x ≠ ± 1. That is the area of \u200b\u200bdefinition of this function (OTZ)
From the roots of the equation x² + 5x + 4 \u003d 0 x1 \u003d -1 x2 \u003d -4, only x \u003d -4 is included in the definition area.
To find the zeros of the function specified graphically, you need to find the intersection points of the function with the abscissa axis.
If the schedule does not cross the Ox axis, the function does not have zeros.
the function whose graph is depicted in the figure, has four zero -
In the algebra, the task of finding zeros of the function is found both in the form of an independent task and when solving other tasks, for example, when studying the function, solving inequalities, etc.
www.algebraclass.ru.
Rule zero function
Basic concepts and properties of functions
Rule (law) compliance. Monotonous function .
Limited and unlimited features. Continuous I.
breaking function . Summer and odd functions.
Periodic function. Period of function.
Zero function . Asymptote .
Definition area and function values. In elementary mathematics, functions are studied only on a plurality of valid numbers. R. . This means that the function argument can only take those actual values \u200b\u200bin which the function is defined, since. It also accepts valid values. Lots of X. all permissible valid values \u200b\u200bof the argument x. in which the function y. = f. (x.) is defined, called function definition area. Lots of Y. All valid values y. which takes a function called function values \u200b\u200barea . Now you can give a more accurate definition of the function: rule (law) conformity between sets X. and Y. , for which for each element from the set X. You can find one and only one element from the set. Y. , called function .
From this definition it follows that the function is considered specified if:
- The function of determining the function X. ;
- the function of the function values \u200b\u200bis specified. Y. ;
- the rule (law) of compliance, and such that for each
the values \u200b\u200bof the argument can be found only one function value.
This requirement of unambiguous function is mandatory.
Monotonous function.
If for any two values \u200b\u200bof the argument x. 1 I. x. 2 from the condition x. 2 > x. 1 follows f. (x. 2) > f. (x. 1) then the function f. (x.) Called increasing; If for any x. 1 I. x. 2 from the condition x. 2 > x. 1 follows f. (x. 2) 
The function depicted in Fig.3 is limited, but not monotonous. The function in Fig. 4 is just the opposite, monotonous, but unlimited. (Explain this, please!).
Continuous and discontinuous function. Function y. = f. (x.) Called continuous at point x. = a. , if a:
1) the function is determined when x. = a. , t. e. f. (a.) exist;
2) exists finite Liming limit f. (x.) ;
If at least one of these conditions is not performed, then the function is called bursting At point x. = a. .
If the function is continuous in all points of their field definitionthen it is called continuous function.
Summer and odd functions. If for anyone x. From the function of determining the function, there is a place: f. (— x.) = f. (x.), then the function is called thought; If there is a place: f. (— x.) = — f. (x.), then the function is called odd. Schedule a function symmetrical relative to the Y axis (Fig. 5), a chart of odd function sim metric relative to the start of coordinates (Fig. 6).
Periodic function. Function f. (x.) — periodic if there is such non-zero number T. that for anyone x. From the function of determining the function, there is a place: f. (x. + T.) = f. (x.). That the fewest The number is called period of function. All trigonometric functions are periodic.
PRI M E P 1. Prove that sin x. It has a period of 2.
R E W E N E. We know that Sin ( x +. 2 n.) \u003d SIN. x. where n. \u003d 0, ± 1, ± 2, ...
Consequently, adding 2 n. to the argument of sinus not
changes his values \u200b\u200bE. Is there another number with such
Let's pretend that P. - such a number, t. E. equality:
fair for any value x. . But then it has
place and primary x. \u003d / 2, t. E.
sin (/ 2 + P.) \u003d SIN / 2 \u003d 1.
But according to the formula of SIN (/ 2 + P.) \u003d COS. P. . Then
of the last two equalities it follows that COS P. \u003d 1, but we
we know that this is true only when P. = 2 n. . Since the smallest
different from zero number of 2 n. is 2, then this is the number
and there is a period of sin x. . Similarly, it is proved that 2
is a period and COS x. .
Prove that Tan functions x. and Cot. x. have a period.
PRI ME R 2. What number is a period of function SIN 2 x. ?
R E W E N E. Consider SIN 2 x. \u003d SIN (2 x +. 2 n.) \u003d sin [2 ( x. + n.) ] .
We see that adding n. To the argument x. does not change
the value of the function. The smallest different number
of n. There is thus this period SIN 2 x. .
Zero function. The value of the argument at which the function is 0, called zero ( root) Functions . The function may have several zeros. For example, a function y. = x. (x. + 1) (x. - 3) has three zero: x. = 0, x. = — 1, x. \u003d 3. Geometrically zero function – this is the abscissa point of intersection of the schedule function with the axis H. .
Figure 7 shows a graph of the function with zeros: x. = a. , x. = b. and x. = c. .
Asymptote. If the graph of the function is unlimited approaching some direct when its removal from the start of the coordinates, then this direct is called asimptoto.
Topic 6. "Interval Method".
If f (x) f (x 0) at x x 0, then the function f (x) is called continuous at point x 0.
If the function is continuous at each point of some gap i, then it is called continuous on the interval I (interval I call interval of continuity function). The graph of the function at this gap is a continuous line, which they say that it can be "drawing, without taking a pencil from paper."
Property of continuous functions.
If, on the interval (a; b), the function F is continuous and does not appeal to zero, then it remains a permanent sign on this interval.
On this property, the method of solving inequalities with one variable is based - the interval method. Suppose that the function f (x) is continuous on the interval I and turns to zero in the end number of points of this interval. By the property of continuous functions, these points I are divided into intervals, in each of which the continuous function f (x) C protects the permanent sign. To determine this sign, it is enough to calculate the value of the function f (x) at any one point from each such interval. Based on this, we obtain the following algorithm for the solution of inequalities by intervals.
Interval method for form inequalities
Interval method. Middle level.
Want to test your strength and find out the result How much are you ready for the exam or OGE?
Linear function
Linear is called the function of the form. Consider the function for example. It is positive at 3 "\u003e and negative when. Point is zero function (). Let's show the signs of this function on the numeric axis:
We say that "the function changes the sign when moving through the point."
It can be seen that the functions of the function correspond to the position of the function of the function: if the schedule is above the axis, the sign "", if below - "".
If we generalize the resulting rule on an arbitrary linear function, we obtain such an algorithm:
Quadratic function
I hope you remember how square inequalities are solved? If not, read the topic "Square inequalities". Let me remind you a general view of a quadratic function :.
Now let's remember which signs receive a quadratic function. Its graph - Parabola, and the function takes the sign "" with such in which Parabola is above the axis, and "" - if Parabola is below the axis:
If the function has zeros (the values \u200b\u200bin which), parabola crosses the axis at two points - the roots of the corresponding square equation. Thus, the axis is divided into three intervals, and the signs of the function alternately change when moving through each root.
Is it possible to somehow define signs without drawing every time a parabola?
Recall that the square three decrease can be decomposed on the factors:
Note roots on the axis:
We remember that the function sign can only change when moving through the root. We use this fact: for each of the three intervals to which the axis is divided with roots, it is enough to determine the function of the function only in one arbitrarily selected point: at the other points of the interval the sign will be the same.
In our example: at 3 "\u003e both expressions in brackets are positive (we substitute, for example: 0"\u003e). We put on the axis sign "":
Well, when (submits, for example,) both brackets are negative, it means that the work is positive:
That's what it is interval method: Knowing the signs of the factors at each interval, we define the sign of all the work.
Consider also cases when there is no zeros of the function, or it is only one.
If there are no, then there are no roots. So, there will be no "transition through the root". So, the function on the entire numeric axis takes only one sign. It is easy to determine, substituting into the function.
If the root is only one, the parabol is touched by the axis, so the function sign does not change when moving through the root. What rule will come up for such situations?
If you decompose such a function on multipliers, two identical multipliers will turn out:
And any expression in the square is nonnegative! Therefore, the function of the function does not change. In such cases, we will allocate the root, when moving through which the sign does not change, circled with a square:
Such root will be called multiple.
Interval method in inequalities
Now any square inequality can be solved without drawing a parabola. It is enough just to place the signs of the quadratic function on the axis, and select the intervals depending on the sign of inequality. For example:
Mind roots on the axis and lay signs:
We need part of the axis with the sign ""; Since the inequality of the unrest, the roots themselves are also included in the solution:
Now consider rational inequality - inequality, both parts of which are rational expressions (see "Rational Equations").
Example:
All factors except one - - here "linear", that is, contain a variable only in the first degree. Such linear multipliers are needed to apply the interval method - a sign when moving through their roots changes. But the multiplier does not have roots at all. This means that it is always positive (check it yourself), and therefore does not affect the sign of all inequality. It means that it can be divided by the left and right-hand side of the inequality, and thus get rid of it:
Now everything is the same as it was with square inequalities: we determine which points each of the multipliers turn to zero, mark these points on the axis and arrange signs. I pay attention very important fact:
In the case of an even number, we do the same as before: we supply the point by the square and do not change the sign when moving through the root. But in the case of an odd amount, this rule is not executed: the sign will still be changed during the transition through the root. Therefore, with such a root, we do not additionally do nothing, as if it is not a multiple. The above rules relate to all even and odd degrees.
What we write in the answer?
If there is a violation of the alternation of signs, it is necessary to be very attentive, because with incomprehensible inequality in response all painted points. But some of Nah often stand a mansion, that is, not included in the painted area. In this case, we add them to the answer as insulated points (in curly brackets):
Examples (Solving yourself):
Answers:
- If among the multipliers is simply a root, because it can be represented as.
.
What is zero functions? Answer is quite simple - this is a mathematical term under which the scope of determining the specified function is implied on which its value is zero. The zeros of the functions are also called the easiest way to explain what zero functions are on several simple examples.
Examples
Consider a simple equation y \u003d x + 3. Since the zero of the function is the value of the argument at which it has acquired zero value, we substitute 0 to the left part of the equation:
In this case, -3 and there is a desired zero. For this function, there is only one root of the equation, but it happens not always.
Consider another example:
Substitute 0 to the left part of the equation, as in the previous example:
Obviously, in this case zeros, the functions will be two: x \u003d 3 and x \u003d -3. If the equation was the argument of the third degree, zeros would be three. It is possible to make a simple conclusion that the number of roots of the polynomial corresponds to the maximum degree of the agructure in the equation. However, many functions, for example y \u003d x 3, at first glance contradicts this statement. Logic and common sense suggest that this function has only one zero at point x \u003d 0. But in fact, the roots are three, they simply all coincide. If you decide the equation in a comprehensive form, it becomes obvious. x \u003d 0 In this case, the root, the multiplicity of which 3. In the previous example, zeros did not coincide, because they had multiplicity 1.
Algorithm Definition
From the examples presented, it can be seen how to determine the zeros of the functions. The algorithm is always the same:
- Write a function.
- Substitute y or f (x) \u003d 0.
- Solve the resulting equation.
The complexity of the last paragraph depends on the degree of the argument of the equation. When solving high degrees equations, it is especially important to remember that the number of roots of the equation is equal to the maximum degree of argument. This is especially important for trigonometric equations, where the division of both parts for sine or cosine leads to the loss of roots.
The equations of random degree is the easiest way to solve the method of the city, which was designed specifically for finding zeros of an arbitrary polynomial.
The value of the functions of functions can be both negative and positive, valid or lying in a complex plane, single or multiple. Or the roots of the equation may not be. For example, the function y \u003d 8 does not acquire zero value with any x, because it does not depend on this variable.
The equation y \u003d x 2 -16 has two roots, and both lie in the complex plane: X 1 \u003d 4I, x 2 \u003d -4I.
Typical errors
A common mistake that schoolchildren admits, not yet impassive in the fact that such a function is, is a replacement for zero argument (x), and not the values \u200b\u200b(y) of the function. They confidently substitute the x \u003d 0 equation and, based on this, find from. But this is the wrong approach.
Another error, as already mentioned, reducing the sine or cosine in the trigonometric equation, due to which one or several zeros of the function is lost. This does not mean that in such equations it is impossible to cut anything, just with further counts it is necessary to take into account these "lost" womb.
Graphic representation
Understand what zeros are function, you can use mathematical programs such as Maple. It can be built in it by specifying the desired number of points and the desired scale. Those points in which the schedule will cross the axis oh, and there are desired zeros. This one of the fastest ways to find the roots of the polynomial, especially if its order is higher than the third. So if there is a need to regularly perform mathematical calculations, find the roots of polynomials of arbitrary degrees, to build graphs, Maple or a similar program will simply be indispensable for the implementation and verification of calculations.
The mathematical representation of the function shows visually how one value entirely determines the value of another value. Traditionally, the numeric functions that are put in touch with the other numbers. Zero functions, usually call the value of the argument at which the function adds to zero.
Instruction
1. In order to detect the zeros of the functions, it is necessary to equate its right-to-zero and solve the obtained equation. Imagine, you are given a function f (x) \u003d x-5.
2. To find the zeros of this function, take and equate its right side to zero: x-5 \u003d 0.
3. After deciding this equation, we obtain that X \u003d 5 and this value of the argument and will be zero functions. That is, with the value of the argument 5, the function f (x) appeals to zero.
Under the submission functions In mathematics, they understand the relationship between elements of sets. If we say more correctly, this "law", along which the whole element of one set (called the definition region) is put in accordance with a certain element of another set (called the area of \u200b\u200bthe values).
You will need
- Knowledge in the field of algebra and mathematical review.
Instruction
1. Values functions This is a certain area, the values \u200b\u200bof which can receive a function. Say the value area functions f (x) \u003d | x | from 0 to infinity. In order to detect value functions At a certain point you need to substitute argument functions its numeric equivalent, the resulting number and will value M. functions . Let the function f (x) \u003d | x | - 10 + 4X. Find value functions At point x \u003d -2. Substitute instead of X Number -2: F (-2) \u003d | -2 | - 10 + 4 * (- 2) \u003d 2 - 10 - 8 \u003d -16. I.e value functions At point -2 is -16.
Note!
Earlier than to look for the value of the function at the point - make sure it enters the function of defining the function.
Useful advice
A similar method is allowed to detect the value of the function of several arguments. The difference is that in return one number will need to substitute several - by the number of arguments of the function.
The function is the installed connection of the variable from the variable x. Moreover, the value of X, called the argument, corresponds to the exceptional value of the function. In graphical form, the function is depicted on a decartular coordinate system in the form of a graph. The intersection points of the graph with the abscissa axis on which the arguments are deposited x are called zeros of the function. Search for permissible zeros is one of the tasks of finding a given function. At the same time, all permissible values \u200b\u200bof the independent variable X forming the function of determining the function (OOOF) are taken into account.
Instruction
1. Zero function is such a value of the arodes of x, at which the function value is zero. However, only the arguments that are included in the field of determining the function under study can be. That is, in such a lot of values \u200b\u200bfor which the function f (x) has a sense.
2. Write down the specified function and equate it to zero, say f (x) \u003d 2x? + 5x + 2 \u003d 0. Decide the resulting equation and detect its valid roots. The roots of the square equation are calculated with the support of the discriminant. 2x? + 5x + 2 \u003d 0; d \u003d b? -4ac \u003d 5? -4 * 2 * 2 \u003d 9; x1 \u003d (-b +? D) / 2 * a \u003d (-5 + 3) / 2 * 2 \u003d -0.5; x2 \u003d (-b-? D) / 2 * a \u003d (-5-3) / 2 * 2 \u003d -2. In this case, two roots of the square equation corresponding to the initial function arguments were obtained f (x).
3. All detected x values \u200b\u200bcheck for belonging to the definition area of \u200b\u200bthe specified function. Discover the OOF, for this, check the initial expression on the presence of an even degree of type roots? F (x), to the presence of fractions in the function with an argument in the denominator, for the presence of logarithmic or trigonometric expressions.
4. Considering the function with the expression under the root of an even degree, take over the definition area all the arguments of which the values \u200b\u200bof which do not turn the feeding expression into a negative number (in front of the function does not make sense). Specify whether the function detected functions fall into a certain area of \u200b\u200bpermissible values \u200b\u200bx.
5. The denomoter of the fraction cannot appeal to zero, will investigantly eliminate the arguments of x that lead to such a result. For logarithmic values, only those values \u200b\u200bof the argument under which the expression itself is much zero. Zero functions that pay for the grocery expression in zero or a negative number must be discarded from the final result.
Note!
By finding the roots of the equation, extra roots may occur. Check it easy: Quite to substitute the obtained value of the argument to the function and make sure whether the function in zero appears.
Useful advice
Occasionally, the function is not expressed in an obvious form through its arguments, then it is easy to know that it represents this feature. An example of this can serve as the equation of a circle.