Even m is an odd function. Function properties. Examining a function for an extremum
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Twitter even if for all \ (x \) from its domain of definition it is true: \ (f (-x) = f (x) \).
The graph of an even function is symmetric about the \ (y \) axis:
Example: the function \ (f (x) = x ^ 2 + \ cos x \) is even, because \ (f (-x) = (- x) ^ 2 + \ cos ((- x)) = x ^ 2 + \ cos x = f (x) \).
\ (\ blacktriangleright \) The \ (f (x) \) function is called odd if for all \ (x \) from its domain it is true: \ (f (-x) = - f (x) \).
The graph of an odd function is symmetric about the origin:
Example: the function \ (f (x) = x ^ 3 + x \) is odd because \ (f (-x) = (- x) ^ 3 + (- x) = - x ^ 3-x = - (x ^ 3 + x) = - f (x) \).
\ (\ blacktriangleright \) Functions that are neither even nor odd are called functions general view... Such a function can always be uniquely represented as a sum of an even and an odd function.
For example, the function \ (f (x) = x ^ 2-x \) is the sum of an even function \ (f_1 = x ^ 2 \) and an odd function \ (f_2 = -x \).
\ (\ blacktriangleright \) Some properties:
1) The product and quotient of two functions of the same parity is an even function.
2) The product and quotient of two functions of different parity is an odd function.
3) The sum and difference of even functions is an even function.
4) The sum and difference of odd functions is an odd function.
5) If \ (f (x) \) is an even function, then the equation \ (f (x) = c \ (c \ in \ mathbb (R) \)) has a unique root if and only if, when \ (x = 0 \).
6) If \ (f (x) \) is an even or odd function, and the equation \ (f (x) = 0 \) has a root \ (x = b \), then this equation will necessarily have a second root \ (x = -b \).
\ (\ blacktriangleright \) A function \ (f (x) \) is called periodic on \ (X \) if \ (f (x) = f (x + T) \), where \ (x, x + T \ in X \). The smallest \ (T \) for which this equality holds is called the main (main) period of the function.
A periodic function has any number of the form \ (nT \), where \ (n \ in \ mathbb (Z) \) will also be a period.
Example: any trigonometric function is periodic;
for the functions \ (f (x) = \ sin x \) and \ (f (x) = \ cos x \), the principal period is \ (2 \ pi \), for the functions \ (f (x) = \ mathrm ( tg) \, x \) and \ (f (x) = \ mathrm (ctg) \, x \) the principal period is \ (\ pi \).
In order to plot a graph of a periodic function, you can plot its graph on any segment of length \ (T \) (main period); then the graph of the entire function is completed by shifting the constructed part by an integer number of periods to the right and left:
\ (\ blacktriangleright \) The domain \ (D (f) \) of a function \ (f (x) \) is a set consisting of all values of the \ (x \) argument for which the function is meaningful (defined).
Example: the function \ (f (x) = \ sqrt x + 1 \) has scope: \ (x \ in
Task 1 # 6364
Task level: Equal to the exam
For what values of the parameter \ (a \) the equation
has the only solution?
Note that since \ (x ^ 2 \) and \ (\ cos x \) are even functions, then if the equation has a root \ (x_0 \), it will also have a root \ (- x_0 \).
Indeed, let \ (x_0 \) be a root, that is, the equality \ (2x_0 ^ 2 + a \ mathrm (tg) \, (\ cos x_0) + a ^ 2 = 0 \) right. Substitute \ (- x_0 \): \ (2 (-x_0) ^ 2 + a \ mathrm (tg) \, (\ cos (-x_0)) + a ^ 2 = 2x_0 ^ 2 + a \ mathrm (tg) \, (\ cos x_0) + a ^ 2 = 0 \).
Thus, if \ (x_0 \ ne 0 \), then the equation will already have at least two roots. Therefore, \ (x_0 = 0 \). Then:
We got two values for the \ (a \) parameter. Note that we have used the fact that \ (x = 0 \) is exactly the root of the original equation. But we have never used the fact that he is the only one. Therefore, it is necessary to substitute the resulting values of the parameter \ (a \) into the original equation and check for which specific \ (a \) the root \ (x = 0 \) will really be unique.
1) If \ (a = 0 \), then the equation takes the form \ (2x ^ 2 = 0 \). Obviously, this equation has only one root \ (x = 0 \). Therefore, the value \ (a = 0 \) suits us.
2) If \ (a = - \ mathrm (tg) \, 1 \), then the equation takes the form \ We rewrite the equation as \ Because \ (- 1 \ leqslant \ cos x \ leqslant 1 \), then \ (- \ mathrm (tg) \, 1 \ leqslant \ mathrm (tg) \, (\ cos x) \ leqslant \ mathrm (tg) \, 1 \)... Therefore, the values of the right-hand side of equation (*) belong to the segment \ ([- \ mathrm (tg) ^ 2 \, 1; \ mathrm (tg) ^ 2 \, 1] \).
Since \ (x ^ 2 \ geqslant 0 \), the left side of the equation (*) is greater than or equal to \ (0+ \ mathrm (tg) ^ 2 \, 1 \).
Thus, equality (*) can only hold when both sides of the equation are \ (\ mathrm (tg) ^ 2 \, 1 \). This means that \ [\ begin (cases) 2x ^ 2 + \ mathrm (tg) ^ 2 \, 1 = \ mathrm (tg) ^ 2 \, 1 \\ \ mathrm (tg) \, 1 \ cdot \ mathrm (tg) \ , (\ cos x) = \ mathrm (tg) ^ 2 \, 1 \ end (cases) \ quad \ Leftrightarrow \ quad \ begin (cases) x = 0 \\ \ mathrm (tg) \, (\ cos x) = \ mathrm (tg) \, 1 \ end (cases) \ quad \ Leftrightarrow \ quad x = 0 \] Therefore, the value \ (a = - \ mathrm (tg) \, 1 \) suits us.
Answer:
\ (a \ in \ (- \ mathrm (tg) \, 1; 0 \) \)
Quest 2 # 3923
Task level: Equal to the exam
Find all values of the parameter \ (a \), for each of which the graph of the function \
symmetrical about the origin.
If the graph of a function is symmetric about the origin, then such a function is odd, that is, \ (f (-x) = - f (x) \) holds for any \ (x \) from the domain of the function. Thus, it is required to find those values of the parameter for which \ (f (-x) = - f (x). \)
\ [\ begin (aligned) & 3 \ mathrm (tg) \, \ left (- \ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a + 3x) 4 = - \ left (3 \ mathrm (tg) \, \ left (\ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a-3x) 4 \ right) \ quad \ Rightarrow \ quad -3 \ mathrm (tg) \ , \ dfrac (ax) 5 + 2 \ sin \ dfrac (8 \ pi a + 3x) 4 = - \ left (3 \ mathrm (tg) \, \ left (\ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a-3x) 4 \ right) \ quad \ Rightarrow \\ \ Rightarrow \ quad & \ sin \ dfrac (8 \ pi a + 3x) 4+ \ sin \ dfrac (8 \ pi a- 3x) 4 = 0 \ quad \ Rightarrow \ quad2 \ sin \ dfrac12 \ left (\ dfrac (8 \ pi a + 3x) 4+ \ dfrac (8 \ pi a-3x) 4 \ right) \ cdot \ cos \ dfrac12 \ left (\ dfrac (8 \ pi a + 3x) 4- \ dfrac (8 \ pi a-3x) 4 \ right) = 0 \ quad \ Rightarrow \ quad \ sin (2 \ pi a) \ cdot \ cos \ frac34 x = 0 \ end (aligned) \]
The last equation must be satisfied for all \ (x \) from the domain \ (f (x) \), therefore, \ (\ sin (2 \ pi a) = 0 \ Rightarrow a = \ dfrac n2, n \ in \ mathbb (Z) \).
Answer:
\ (\ dfrac n2, n \ in \ mathbb (Z) \)
Quest 3 # 3069
Task level: Equal to the exam
Find all values of the parameter \ (a \), for each of which the equation \ has 4 solutions, where \ (f \) is an even periodic function with a period \ (T = \ dfrac (16) 3 \) defined on the whole number line , and \ (f (x) = ax ^ 2 \) for \ (0 \ leqslant x \ leqslant \ dfrac83. \)
(Task from subscribers)
Since \ (f (x) \) is an even function, its graph is symmetric about the ordinate axis, therefore, for \ (- \ dfrac83 \ leqslant x \ leqslant 0 \)\ (f (x) = ax ^ 2 \). Thus, for \ (- \ dfrac83 \ leqslant x \ leqslant \ dfrac83 \), and this is a segment of length \ (\ dfrac (16) 3 \), function \ (f (x) = ax ^ 2 \).
1) Let \ (a> 0 \). Then the graph of the function \ (f (x) \) will look like this: 
Then, in order for the equation to have 4 solutions, it is necessary that the graph \ (g (x) = | a + 2 | \ cdot \ sqrtx \) passes through the point \ (A \): 
Hence, \ [\ dfrac (64) 9a = | a + 2 | \ cdot \ sqrt8 \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & 9 (a + 2) = 32a \\ & 9 (a +2) = - 32a \ end (aligned) \ end (gathered) \ right. \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & a = \ dfrac (18) (23) \\ & a = - \ dfrac (18) (41) \ end (aligned) \ end ( gathered) \ right. \] Since \ (a> 0 \), then \ (a = \ dfrac (18) (23) \) is suitable.
2) Let \ (a<0\)
. Тогда картинка окажется симметричной относительно начала координат:
It is necessary that the graph \ (g (x) \) passes through the point \ (B \): \ [\ dfrac (64) 9a = | a + 2 | \ cdot \ sqrt (-8) \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & a = \ dfrac (18) (23 ) \\ & a = - \ dfrac (18) (41) \ end (aligned) \ end (gathered) \ right. \] Since \ (a<0\)
, то подходит \(a=-\dfrac{18}{41}\)
.
3) The case when \ (a = 0 \) does not fit, since then \ (f (x) = 0 \) for all \ (x \), \ (g (x) = 2 \ sqrtx \) and the equation will only have 1 root.
Answer:
\ (a \ in \ left \ (- \ dfrac (18) (41); \ dfrac (18) (23) \ right \) \)
Quest 4 # 3072
Task level: Equal to the exam
Find all values \ (a \), for each of which the equation \
has at least one root.
(Task from subscribers)
We rewrite the equation as \
and consider two functions: \ (g (x) = 7 \ sqrt (2x ^ 2 + 49) \) and \ (f (x) = 3 | x-7a | -6 | x | -a ^ 2 + 7a \ ).
The function \ (g (x) \) is even, has a minimum point \ (x = 0 \) (moreover, \ (g (0) = 49 \)).
The function \ (f (x) \) for \ (x> 0 \) is decreasing, and for \ (x<0\)
– возрастающей, следовательно, \(x=0\)
– точка максимума.
Indeed, for \ (x> 0 \) the second module expands positively (\ (| x | = x \)), therefore, regardless of how the first module expands, \ (f (x) \) will be equal to \ ( kx + A \), where \ (A \) is an expression from \ (a \), and \ (k \) is either \ (- 9 \) or \ (- 3 \). For \ (x<0\)
наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\)
, где \(k\)
равно либо \(3\)
, либо \(9\)
.
Find the value \ (f \) at the maximum point: \ 
In order for the equation to have at least one solution, the graphs of the functions \ (f \) and \ (g \) must have at least one intersection point. Therefore, you need: \ \\]
Answer:
\ (a \ in \ (- 7 \) \ cup \)
Task 5 # 3912
Task level: Equal to the exam
Find all values of the parameter \ (a \), for each of which the equation \
has six different solutions.
Let's make the replacement \ ((\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) = t \), \ (t> 0 \). Then the equation takes the form \
We will gradually write down the conditions under which the original equation will have six solutions.
Note that the quadratic equation \ ((*) \) can have at most two solutions. Any cubic equation \ (Ax ^ 3 + Bx ^ 2 + Cx + D = 0 \) can have at most three solutions. Therefore, if the equation \ ((*) \) has two different solutions (positive !, since \ (t \) must be greater than zero) \ (t_1 \) and \ (t_2 \), then, having made the reverse substitution, we we get: \ [\ left [\ begin (gathered) \ begin (aligned) & (\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) = t_1 \\ & (\ sqrt2) ^ (x ^ 3-3x ^ 2 +4) = t_2 \ end (aligned) \ end (gathered) \ right. \] Since any positive number can be represented as \ (\ sqrt2 \) to some extent, for example, \ (t_1 = (\ sqrt2) ^ (\ log _ (\ sqrt2) t_1) \), then the first equation of the set will be rewritten as \
As we have already said, any cubic equation has at most three solutions, therefore, each equation from the set will have at most three solutions. This means that the entire set will have no more than six solutions.
This means that for the original equation to have six solutions, the quadratic equation \ ((*) \) must have two different solutions, and each obtained cubic equation (from the set) must have three different solutions (and no solution of one equation must coincide with which one - or by the decision of the second!)
Obviously, if the quadratic equation \ ((*) \) has one solution, then we will not get six solutions of the original equation.
Thus, the solution plan becomes clear. Let's write down the conditions that must be met, point by point.
1) For the equation \ ((*) \) to have two different solutions, its discriminant must be positive: \
2) You also need both roots to be positive (since \ (t> 0 \)). If the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Therefore, you need: \ [\ begin (cases) 12-a> 0 \\ - (a-10)> 0 \ end (cases) \ quad \ Leftrightarrow \ quad a<10\]
Thus, we have already provided ourselves with two different positive roots \ (t_1 \) and \ (t_2 \).
3)
Let's take a look at an equation like this \
For which \ (t \) will it have three different solutions? Thus, we have determined that both roots of the equation \ ((*) \) must lie in the interval \ ((1; 4) \). How do you write this condition? had four different nonzero roots representing, together with \ (x = 0 \), an arithmetic progression. Note that the function \ (y = 25x ^ 4 + 25 (a-1) x ^ 2-4 (a-7) \) is even, so if \ (x_0 \) is the root of the equation \ ((*) \ ), then \ (- x_0 \) will also be its root. Then it is necessary that the roots of this equation are numbers ordered in ascending order: \ (- 2d, -d, d, 2d \) (then \ (d> 0 \)). It is then that these five numbers will form an arithmetic progression (with the difference \ (d \)). For these roots to be the numbers \ (- 2d, -d, d, 2d \), it is necessary that the numbers \ (d ^ (\, 2), 4d ^ (\, 2) \) be the roots of the equation \ (25t ^ 2 +25 (a-1) t-4 (a-7) = 0 \). Then by Vieta's theorem: We rewrite the equation as \
and consider two functions: \ (g (x) = 20a-a ^ 2-2 ^ (x ^ 2 + 2) \) and \ (f (x) = 13 | x | -2 | 5x + 12a | \) ... In order for the equation to have at least one solution, the graphs of the functions \ (f \) and \ (g \) must have at least one intersection point. Therefore, you need: \
Solving this set of systems, we get the answer: \\]
Answer: \ (a \ in \ (- 2 \) \ cup \) Check if the graph of the function is symmetrical about the y-axis. Symmetry refers to the mirroring of the chart about the ordinate axis. If the portion of the graph to the right of the y-axis (positive explanatory variable) coincides with the portion of the graph to the left of the y-axis (negative explanatory variable), the graph is symmetrical about the y-axis. If the function is symmetric about the ordinate, the function is even. Check if the graph of the function is symmetrical about the origin. The origin is the point with coordinates (0,0). Symmetry about the origin means that a positive value y (\ displaystyle y)(with a positive value x (\ displaystyle x)) corresponds to a negative value y (\ displaystyle y)(with a negative value x (\ displaystyle x)), and vice versa. Odd functions are symmetric about the origin. Check if the graph of the function has any symmetry. The last type of function is a function whose graph does not have symmetry, that is, there is no mirroring both about the ordinate axis and about the origin. For example, given a function. Attention! Slide previews are for informational purposes only and may not represent all presentation options. If you are interested in this work, please download the full version. Goals: Equipment: multimedia installation, interactive whiteboard, handouts. Forms of work: frontal and group with elements of search and research activities. Information sources: DURING THE CLASSES 1. Organizational moment Setting the goals and objectives of the lesson. 2.
Homework check No. 10.17 (Problem book 9kl. A. G. Mordkovich). a) at = f(NS), f(NS) = b) f (–2) = –3; f (0) = –1; f(5) = 69; c) 1. D ( f) = [– 2; + ∞) (Did you use the function research algorithm?) Slide.
2. Let's check the table that you were asked on the slide. Domain Function zeros Intervals of constancy x = –5, x € (–5; 3) U х € (–∞; –5) U x ∞ –5, x € (–5; 3) U х € (–∞; –5) U x ≠ –5, х € (–∞; –5) U x € (–5; 2) 3.
Knowledge update - Given functions. NS ≠ 0 and not defined. 4. New material
- In carrying out this work, guys, we identified one more property of a function that is unfamiliar to you, but no less important than the others - this is the even and odd function. Write down the topic of the lesson: "Even and odd functions", our task is to learn how to determine the evenness and oddness of a function, to find out the significance of this property in the study of functions and plotting. Def. 1 Function at = f (NS) given on the set X is called even if for any value NSЄ X is executed equality f (–x) = f (x). Give examples. Def. 2 Function y = f (x) given on the set X is called odd if for any value NSЄ X the equality f (–x) = –f (x) holds. Give examples. Where have we encountered the terms "even" and "odd"? The study of the question of whether a function is even or odd is called the study of a function for parity. Slide Definitions 1 and 2 dealt with the values of the function for x and - x, thus it is assumed that the function is also defined for the value NS, and at - NS. Def 3. If number set together with each of its elements x contains the opposite element -x, then the set NS called a symmetric set. Examples: (–2; 2), [–5; 5]; (∞; ∞) are symmetric sets, and, [–5; 4] are asymmetric. - Is the domain of definition of even functions a symmetric set? The odd ones? Slide
Algorithm for analyzing a function for parity 1. Determine whether the function domain is symmetric. If not, then the function is neither even nor odd. If yes, then go to step 2 of the algorithm. 2. Write an expression for f(–NS). 3. Compare f(–NS).and f(NS): Examples:
Investigate the function for parity a) at= x 5 +; b) at=; v) at= . Solution. a) h (x) = x 5 +, 1) D (h) = (–∞; 0) U (0; + ∞), symmetric set. 2) h (- x) = (–x) 5 + - x5 - = - (x 5 +), 3) h (- x) = - h (x) => function h (x)= x 5 + odd. b) y =, at = f(NS), D (f) = (–∞; –9)? (–9; + ∞), asymmetric set, so the function is neither even nor odd. v) f(NS) =, y = f (x), 1) D ( f) = (–∞; 3] ≠; b) (∞; –2), (–4; 4]? Option 2 1. Is the given set symmetric: a) [–2; 2]; b) (∞; 0], (0; 7)? a) y = x 2 (2x - x 3), b) y = Mutual verification of slide. 6. Assignment at home: №11.11, 11.21,11.22; Proof of the geometric meaning of the parity property. *** (Setting the USE option). 1. The odd function y = f (x) is defined on the whole number line. For any non-negative value of the variable x, the value of this function coincides with the value of the function g ( NS)
= NS(NS + 1)(NS + 3)(NS- 7). Find the value of the function h ( NS) = for NS = 3. 7. Summing up Function is one of the most important mathematical concepts. Function - Variable Dependency at from variable x if each value NS matches a single value at... Variable NS called the independent variable or argument. Variable at called the dependent variable. All values of the independent variable (variable x) form the domain of the function. All values that the dependent variable (variable y), form the range of values of the function. Function graph call the set of all points coordinate plane, the abscissas of which are equal to the values of the argument, and the ordinates are equal to the corresponding values of the function, that is, the values of the variable are plotted along the abscissa axis x, and the ordinate represents the values of the variable y... To plot a function graph, you need to know the properties of the function. The main properties of the function will be discussed later! To graph a function, we recommend using our program - Graphing Functions Online. If you have any questions while 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! Basic properties of functions. 1) Function domain and function domain. Function scope is the set of all valid valid values of the argument x(variable x) for which the function y = f (x) defined. In elementary mathematics, functions are studied only on the set of real numbers. 2) Function zeros. The values NS at which y = 0 is called function zeros... These are the abscissas of the points of intersection of the graph of the function with the Ox axis. 3) Intervals of constancy of function. Intervals of constant sign of function - such intervals of values x, on which the values of the function y either only positive, or only negative, are called intervals of constancy of the function. 4) Monotonicity of function. An increasing function (in a certain interval) is a function for which a larger value of the argument from this interval corresponds to a larger value of the function. Decreasing function (in a certain interval) - a function for which the larger value of the argument from this interval corresponds to the smaller value of the function. 5) Parity (odd) function. An even function is a function whose domain of definition is symmetric about the origin and for any NS f (-x) = f (x)... The graph of an even function is symmetric about the ordinate axis. An odd function is a function whose domain of definition is symmetric about the origin and for any NS from the domain of definition, the equality f (-x) = - f (x). The graph of an odd function is symmetric about the origin. Even function Odd function has the following properties: Not every function is odd or even. Functions general view are neither even nor odd. 6) Limited and unlimited functions. A function is called bounded if there is a positive number M such that | f (x) | ≤ M for all values of x. If there is no such number, then the function is unlimited. 7) Periodicity of the function. A function f (x) is periodic if there is a nonzero number T such that for any x from the domain of the function the following holds: f (x + T) = f (x). This smallest number is called the period of the function. Everything trigonometric functions are periodic. (Trigonometric formulas). Function f is called periodic if there is a number such that for any x from the domain, the equality f (x) = f (x-T) = f (x + T). T is the period of the function. Any periodic function has an infinite set of periods. In practice, the shortest positive period is usually considered. The values of the periodic function are repeated after an interval equal to the period. This is used when building graphs. - (mat.) A function y = f (x) is called even if it does not change when the independent variable only changes sign, that is, if f (x) = f (x). If f (x) = f (x), then the function f (x) is called odd. For example, y = cosx, y = x2 ... ... F (x) = x is an example of an odd function. f (x) = x2 is an example of an even function. f (x) = x3 ... Wikipedia A function satisfying the equality f (x) = f (x). See Even and Odd Functions ... Great Soviet Encyclopedia F (x) = x is an example of an odd function. f (x) = x2 is an example of an even function. f (x) = x3 ... Wikipedia F (x) = x is an example of an odd function. f (x) = x2 is an example of an even function. f (x) = x3 ... Wikipedia F (x) = x is an example of an odd function. f (x) = x2 is an example of an even function. f (x) = x3 ... Wikipedia F (x) = x is an example of an odd function. f (x) = x2 is an example of an even function. f (x) = x3 ... Wikipedia Special functions introduced by the French mathematician E. Mathieu in 1868 when solving problems on the oscillation of an elliptical membrane. M. f. are also used to study the propagation of electromagnetic waves in an elliptical cylinder ... Great Soviet Encyclopedia The "sin" request is redirected here; see also other meanings. The "sec" request is redirected here; see also other meanings. The Sinus request is redirected here; see also other meanings ... Wikipedia
Consider the function \ (f (x) = x ^ 3-3x ^ 2 + 4 \).
Can be factorized: \
Therefore, its zeros are \ (x = -1; 2 \).
If we find the derivative \ (f "(x) = 3x ^ 2-6x \), then we get two extremum points \ (x_ (max) = 0, x_ (min) = 2 \).
Hence, the graph looks like this: 
We see that any horizontal line \ (y = k \), where \ (0
Thus, you need: \ [\ begin (cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\]
Let's also immediately notice that if the numbers \ (t_1 \) and \ (t_2 \) are different, then the numbers \ (\ log _ (\ sqrt2) t_1 \) and \ (\ log _ (\ sqrt2) t_2 \) will be different, hence, the equations \ (x ^ 3-3x ^ 2 + 4 = \ log _ (\ sqrt2) t_1 \) and \ (x ^ 3-3x ^ 2 + 4 = \ log _ (\ sqrt2) t_2 \) will have mismatched roots.
The \ ((**) \) system can be rewritten as follows: \ [\ begin (cases) 1
We will not write out the roots explicitly.
Consider the function \ (g (t) = t ^ 2 + (a-10) t + 12-a \). Its graph is a parabola with upward branches, which has two points of intersection with the abscissa axis (we wrote this condition in point 1)). How should its graph look like so that the points of intersection with the abscissa axis are in the interval \ ((1; 4) \)? So: 
First, the values \ (g (1) \) and \ (g (4) \) of the function at the points \ (1 \) and \ (4 \) must be positive, and secondly, the vertex of the parabola \ (t_0 \ ) must also be in the range \ ((1; 4) \). Therefore, we can write the system: \ [\ begin (cases) 1 + a-10 + 12-a> 0 \\ 4 ^ 2 + (a-10) \ cdot 4 + 12-a> 0 \\ 1<\dfrac{-(a-10)}2<4\end{cases}\quad\Leftrightarrow\quad 4
The function \ (g (x) \) has a maximum point \ (x = 0 \) (moreover, \ (g _ (\ text (vert)) = g (0) = - a ^ 2 + 20a-4 \)):
\ (g "(x) = - 2 ^ (x ^ 2 + 2) \ cdot \ ln 2 \ cdot 2x \)... Derivative zero: \ (x = 0 \). For \ (x<0\)
имеем: \(g">0 \), for \ (x> 0 \): \ (g "<0\)
.
The function \ (f (x) \) for \ (x> 0 \) is increasing, and for \ (x<0\)
– убывающей, следовательно, \(x=0\)
– точка минимума.
Indeed, for \ (x> 0 \) the first module will open positively (\ (| x | = x \)), therefore, regardless of how the second module will open, \ (f (x) \) will be equal to \ ( kx + A \), where \ (A \) is an expression from \ (a \), and \ (k \) is equal to either \ (13-10 = 3 \) or \ (13 + 10 = 23 \). For \ (x<0\)
наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\)
, где \(k\)
равно либо \(-3\)
, либо \(-23\)
.
Find the value \ (f \) at the minimum point: \
Back forward
1.Algebra9class A.G. Mordkovich. Textbook.
2.Algebra grade 9 A.G. Mordkovich. Problem book.
3.Algebra grade 9. Assignments for student learning and development. Belenkova E.Yu. Lebedintseva E.A.
2. E ( f) = [– 3; + ∞)
3. f(NS) = 0 for NS ~ 0,4
4. f(NS)> 0 for NS > 0,4 ; f(NS)
< 0 при – 2 <
NS <
0,4.
5. The function increases with NS € [– 2; + ∞)
6. The function is limited from below.
7. at naim = - 3, at naib does not exist
8. The function is continuous.Fill the table
Coordinates of points of intersection of the graph with Oy
x = 2
U (2; ∞)
U (–3; 2)
x ≠ 2
U (2; ∞)
U (–3; 2)
x ≠ 2
U (2; ∞)
- Specify the scope for each function.
- Compare the value of each function for each pair of argument values: 1 and - 1; 2 and - 2.
- For which of these functions in the domain of definition are the equalities satisfied? f(– NS)
= f(NS), f(– NS) = – f(NS)? (enter the obtained data into the table) Slide
f(1) and f(– 1)
f(2) and f(– 2)
charts
f(– NS) = –f(NS)
f(– NS) = f(NS)
1. f(NS) =
2. f(NS) = NS 3
3. f(NS) = | NS |
4.f(NS) = 2NS – 3
5. f(NS) =
6. f(NS)=
NS >
–1
So, let's find the definitions in the textbook and read (p. 110) ... Slide
Which of these functions do you think will be even? Why? What are odd? Why?
For any function of the form at= x n, where n- an integer it can be argued that the function is odd for n- odd and the function is even for n- even.
- View functions at= and at = 2NS- 3 are neither even nor odd, since equalities are not satisfied f(– NS) = – f(NS), f(–
NS) = f(NS)
- If D ( f) Is an asymmetric set, then what function?
- Thus, if the function at = f(NS) Is even or odd, then its domain of definition D ( f) Is a symmetric set. Is the converse true, if the domain of a function is a symmetric set, then it is even or odd?
- This means that the presence of a symmetric set of domains of definition is a necessary condition, but not sufficient.
- So how do you investigate a function for parity? Let's try to compose an algorithm.
a); b) y = x · (5 - x 2).2. Investigate the function for parity:
3. In fig. plotted at = f(NS), for all NS satisfying the condition NS?
0.
Plot a function graph at = f(NS), if at = f(NS) Is an even function. 
3. In fig. plotted at = f(NS), for all x satisfying the condition x? 0.
Plot a function graph at = f(NS), if at = f(NS) Is an odd function. 
The range of values of a function is the set of all real values y that the function accepts.
1) The domain of definition is symmetric about the point (0; 0), that is, if the point a belongs to the domain of definition, then the point -a also belongs to the domain of definition.
2) For any value x f (-x) = f (x)
3) The graph of an even function is symmetric about the Oy axis.
1) The domain is symmetric about the point (0; 0).
2) for any value x belonging to the domain, the equality f (-x) = - f (x)
3) The graph of an odd function is symmetric about the origin (0; 0).
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