"The woman is created for a man, not a man for a woman" - such a postulate ...
Numerical set X. Consider symmetric relatively zero if for any x.Є X. value - h. Also belongs to the set X..
Function y. = f.(h. X., it is believed even X. x.Є X., f.(h.) = f.(-h.).
At the even function, the graph is symmetrical with respect to the OU axis.
Function y. = f.(h.), which is set on the set X., it is believed oddif the following conditions are satisfied: a) set X. symmetrically relative to zero; b) for any x.Є X., f.(h.) = -f.(-h.).
An odd function has a graph symmetrical on the origin of the coordinates.
Function w. = f.(x.), x.Є X., called periodic on the X.if there is a number T. (T. ≠ 0) (period Functions) that the following conditions are followed:
- h. - T. and h. + T. From set X. for anyone h.Є X.;
- for anyone h.Є X., f.(h. + T.) = f.(h. - T.) = f.(x).
In the case when T. - this is a function period, then any number of species mT.where m.Є Z., m. ≠ 0, it is also the period of this function. The smallest of the positive periods of this function (if it exists) is called its main period.
In the case when T. - the main period of the function, then to build its graph, you can build part of the graph on any of the intervals of the length of length T.and then make a parallel transfer of this plot of graphics along the axis about h. by ± T., ± 2. T., ....
Function y. = f.(h.), limited from below On the set H. BUTwhat for any h.Є X., BUT ≤ f.(h.). Schedule a function that is limited from the bottom on the set X., fully located above straight w. = BUT (This is a horizontal straight line).
Function w. = f.(x.), limited from above On the set H. (it should be defined on this set), if there is a number INwhat for any h.Є X., f.(h.) ≤ IN. The graph of the function that is limited from above on the set X is fully located below the direct w. = IN (This is a horizontal line).
Function, believed limited On the set H. (It should be defined on this set), if it is limited to this set on top and bottom, i.e. there are such numbers BUT and INwhat for any h.Є X. Inequality are performed A. ≤ f.(x.) ≤ B.. Schedule a function that is limited to the set X., fully located in the interval between straight w. = BUT and w. = IN (These are horizontal direct).
Function w. = f. (h.) is considered limited on the set H. (it should be defined on this set) if there is a number FROM \u003e 0, what for any x.Є X., │f.(h.)│≤ FROM.
Function w. = f.(h.), h.Є X., called increasing (inconsiderable) on subset M.FROM X.when for every h. 1 I. h. 2 is M. such that h. 1 < h. 2, fair f.(h. 1) < f.(h. 2) (f.(h. 1) ≤ f.(h. 2)). Or the function y is called increasing On the set TOIf the greater value of the argument from this set corresponds to the greater function value.
Function w. = f.(h.), h.Єx, called descending (non-pulmonary) on subset M.FROM X.when for every h. 1 I. h. 2 is M. such that h. 1 < h. 2, fair f.(h. 1) > f.(h. 2) (f.(h. 1) ≥ f.(h. 2)). Or function w. called decreasing on the set TOIf the greater value of the argument from this set corresponds to the smaller value of the function.
Function w. = f.(x.), h.Є X., called monotonna on subset M.FROM X.if it is a decreasing (non-gaining) or increasing (inconsider) on M..
If the function w. = f.(h.), h.Є X.is descending or increasingly on a subset M.FROM X.then such a function is called strictly monotonous On the set M..
Number M. Call the greatest function value U on the set TOif this number is the value of the function at a certain value x 0 Argument from setTO, and with other values \u200b\u200bof the argument from the set to the function value, no moreM..
Number m. Call the smallest meaning functions y on the set TOif this number is the value of the function at a certain value. h. 0 argument from set TO, and with other values \u200b\u200bof the argument x from the set TOvalues \u200b\u200bof function at no less m..
The main properties of the function With which it is better to start studying and studying this area of \u200b\u200bits definition and meaning. To remember how the graphics of elementary functions are depicted. Only then you can move to the construction of more complex charts. The theme "functions" has broad applications in the economy and other areas of knowledge. Functions are studied throughout the entire course of mathematics and continue to study inhigh schools . There, functions are investigated using the first and second derivatives.
Required and sufficient condition monotonicity of the function in the interval.
Required and sufficient constancy condition of the function in the interval
Theorem
Suppose that the function f (x) is determined in the gap of X and has a finite derivative of F / (X) inside it, and at the ends (if they belong to x) retains continuity. In order to f (x) in X permanent, sufficient condition f / (x) \u003d 0 inside X.
Evidence
Let this condition are fulfilled. Fix some point x0 from the interval X and take any other point x. For the interval [x0, x] or [x, x0], all conditions are satisfied. lagrange Theorems. Therefore, we can write
f (x) -f (x0) \u003d f / (c) (x - x0),
Where C is contained between X0 and X, which means that it is obviously lies inside X. But, by assumption, f / (c) \u003d 0, so that for all x from x
f (x) \u003d f (x0) \u003d const.
Theorem is proved.
Note that the condition expressed is obviously necessary for the constancy of the function.
Corollary. Let the two functions f (x) and g (x) are defined in the gap of X and inside it have finite derivatives f / (x) and g / (x), and at the ends (if they belong to x) retain continuity. If with this f / (x) \u003d g / (x) inside x,
that in the whole range x these functions differ only to the permanent:
f (x) \u003d g (x) + c (c \u003d const).
To prove, it is enough to apply the theorem to the difference F (x) -g (x), since its derivative f / (x) -g / (x) inside X is reduced to zero, then the difference itself in X will be constant.
Theorem (sufficient condition)
If Function F (X) differentialon (a, b) and f / (x) ≥0 (f / (x) ≤0) on (a, b), then f (x) does not decrease (does not increase) on (a, b).
Evidence f (x2) -f (x1) \u003d f / (c) (x2-x1), where c∈ (x1, x2) and the right side is greater than zero, it means f (x2) -f (x1) ≥0 or f ( x2) ≥F (x1) at x2\u003e x1, the function does not decrease. Theorem is proved. Comment If you require that f / (x)\u003e 0 (f / (x)<0), тогда функция строго возрастает (убывает). 6. Required extremum condition. Required sign of existence of extremum: To find the extremums of the function z \u003d f (x, y), you first need to find stationary points of this function, in which the private derivatives z \u003d f (x, y) are zero. To do this, solve the system of equations: The function may have an extremum also at those points where at least one of the partial derivatives does not exist. Condition (1) is a prerequisite for extremum, but it is not sufficient, i.e. In the stationary point of extremum may not be. Consider a sufficient condition of Extremum. Let the point M 0 be a stationary point of the function z \u003d f (x, y), which has continuous private derivatives of the second order at some neighborhood of the point M0, If D\u003e 0, then the extremum at the point M0 is, and the M0 is the point of the minimum at a\u003e 0 and m0 - the maximum point at a<0. Если D<0, то экстремума в точке M0 нет. For d \u003d 0, additional research is required in the neighborhood of the M0 point, we will not consider this case. 7. A sufficient extremma condition. Look in the 6 question. Direction of convexity of the graphics of the function. Points of inflection Let us give the definition of the edge of the graph of the function. Suppose that the function is differentiable on the interval. This means (see §3) that at this interval a function schedule has a tangential, not parallel axis ordinate. Definition. It is said that the function of the function has a bulge down (up) on the interval if the graph of this function within this interval lies above (below) any of its tangential. The following theorem establishes the relationship between the direction of convexity of the function of the function and the sign of its second derivative. This theorem is provided here without proof. Theorem 25.1. Suppose the function has a second derivative on the interval. Then, if this derivative is positive (negative) everywhere in this interval, then the function schedule has a convex in the interval directed down (up). Let's give the deflection point. Suppose that the function is differentiable at the interval, i.e. At any point, the abscissa of which belongs to the interval, the graph of this function has a tangent. Definition. The function of the function of the function is called the inflection point of this graph, if there is such a neighborhood of the point of the abscissa axis, within which the graph of the function of the left and right of the point has different directions of convexity. The graph of the function depicted in Figure 6, on the interval has a convexity aimed at the interval - convexity directional; The point (0,0) is the point of the inflection of this graph. We formulate without proof the necessary condition for the inflection of the graphics of a function having a second derivative. Theorem 25.2. If the function has a second derivative at a point and the graph of this feature has a blister at the point, then. It is clear that the inflection should be searched only at those points of the abscissa axis, in which the function itself is differentiable, and the second derivative of this function is either equal to zero or does not exist. Such points are called critical points of the second kind. Note that the equality zero of the second derivative is necessary, but not sufficient condition of the inflection. For example, the function at the point does not have an inflection, although the second derivative of this function is equal to zero at the point. Theorem 25.3. Suppose the function has a second derivative in some neighborhood of the point, while the point itself is a critical point of the second kind. Then, if within the specified neighborhood, the second derivative has different signs on the left and to the right of the point, then the graph of this function has bendes at the point. increasing At the interval \\ (x \\), if for any \\ (x_1, x_2 \\ in x \\), such that \\ (x_1 The function is called unlawful \\ (\\ BLACKTRIANGLERIGHT \\) The function \\ (f (x) \\) is called descending At the interval \\ (x \\), if for any \\ (x_1, x_2 \\ in x \\), such that \\ (x_1 The function is called non-head At the interval \\ (x \\), if for any \\ (x_1, x_2 \\ in x \\), such that \\ (x_1 \\ (\\ blacktriangleright \\) increasing and decreasing functions are called strictly monotonous, and non-careful and unlawful - just monotonous. \\ (\\ blacktriangleright \\) Basic properties: I. If the function \\ (F (x) \\) is a strictly monotonne to \\ (x \\), then from the equality \\ (x_1 \u003d x_2 \\) (\\ (x_1, x_2 \\ in x \\)) follows \\ (F (x_1) \u003d f (x_2) \\), and vice versa. Example: The function \\ (f (x) \u003d \\ sqrt x \\) is strictly increasing at all \\ (x \\ in \\), so the equation \\ (x ^ 2 \u003d 9 \\) has no more than one solution at this gap, or rather One: \\ (x \u003d -3 \\). the function \\ (F (X) \u003d - \\ DFRAC 1 (X + 1) \\) is strictly increasing at all \\ (x \\ in (-1; + \\ infty) \\), so the equation \\ (- \\ dfrac 1 (x +1) \u003d 0 \\) has no more than one solution at this gap, or rather not one, because The left side numerator can never be zero. III. If the function \\ (f (x) \\) is unreigable (irrepreneurs) and continuously on the segment \\ (\\), and at the ends of the segment it takes the values \u200b\u200b\\ (f (a) \u003d a, f (b) \u003d b \\), then When \\ (C \\ In \\) (\\ (C \\ In \\)), the equation \\ (F (x) \u003d C \\) always has at least one solution. Example: The function \\ (f (x) \u003d x ^ 3 \\) is strictly increasing (that is, strictly monotone) and continuous at all \\ (x \\ in \\ mathbb (r) \\), so at any \\ (C \\ in - \\ infty; + \\ infty) \\) The equation \\ (x ^ 3 \u003d C \\) has exactly one solution: \\ (x \u003d \\ sqrt (c) \\). Task 1 # 3153 Task level: easier to ege it has exactly two roots. Refer to the equation in the form: \\ [(3x ^ 2) ^ 3 + 3x ^ 2 \u003d (x-a) ^ 3 + (x-a) \\] Consider the function \\ (F (T) \u003d T ^ 3 + T \\). Then the equation rewrite in the form: \\ we investigate the function \\ (F (T) \\). It is therefore, the function \\ (f (t) \\) increases at all \\ (t \\). It means that each value of the function \\ (F (T) \\) corresponds to exactly one value of the argument \\ (t \\). Therefore, in order for the equation to have roots, you need: \
In order for the obtained equation to have two roots, it is necessary that its discriminant is positive: \
Answer: \\ (\\ left (- \\ infty; \\ dfrac1 (12) \\ right) \\) Task 2 # 2653 Task level: equal to ege Find all the parameter values \u200b\u200b\\ (a \\), in which the equation \
it has two roots. (Task from subscribers.) We will replace: \\ (ax ^ 2-2x \u003d t \\), \\ (x ^ 2-1 \u003d u \\). Then the equation will take the form: \
Consider the function \\ (F (W) \u003d 7 ^ W + \\ SQRTW \\). Then our equation will take the form: \\ Find a derivative \
Note that at all \\ (w \\ ne 0 \\), the derivative \\ (F "(W)\u003e 0 \\), because \\ (7 ^ w\u003e 0 \\), \\ (W ^ 6\u003e 0 \\). We also note that the function \\ (F (W) \\) is defined at all \\ (W \\). Because \\ (F (W) \\) is continuous, then we can conclude that \\ (F (w) \\) increases on everything \\ (\\ mathbb (R) \\). \
In order for this equation to have two roots, it must be square and its discriminant must be positive: \\ [\\ begin (Cases) A-1 \\ NE 0 \\\\ 4-4 (A-1)\u003e 0 \\ End (Cases) \\ quad \\ leftrightarrow \\ quad \\ begin (Cases) A \u200b\u200b\\ NE1 \\\\ A<2\end{cases}\]
Answer: \\ ((- \\ INFTY; 1) \\ CUP (1; 2) \\) Task 3 # 3921 Task level: equal to ege Find all the positive values \u200b\u200bof the parameter \\ (A \\), in which the equation it has at least \\ (2 \\) solutions. We transfer all the terms comprising \\ (ax \\) to the left, and containing \\ (x ^ 2 \\) - to the right, and consider the function Then the initial equation will take the form: Find a derivative: Because \\ ((T-2) ^ 2 \\ geqslant 0, \\ E ^ T\u003e 0, \\ 1+ \\ COS (2T) \\ GEQSLANT 0 \\), then \\ (f "(t) \\ geqslant 0 \\) for any \\ (t \\ in \\ mathbb (r) \\). Moreover, \\ (f "(t) \u003d 0 \\), if \\ ((t-2) ^ 2 \u003d 0 \\) and \\ (1+ \\ cos (2t) \u003d 0 \\) at the same time, which is not satisfied with any Thus, the function \\ (f (t) \\) strictly increases at all \\ (t \\ in \\ mathbb (r) \\). It means that the equation \\ (F (ax) \u003d f (x ^ 2) \\) is equivalent to the equation \\ (ax \u003d x ^ 2 \\). Equation \\ (x ^ 2-Ax \u003d 0 \\) with \\ (a \u003d 0 \\) has one root \\ (x \u003d 0 \\), and with \\ (a \\ ne 0 \\) has two different roots \\ (x_1 \u003d 0 \\) and \\ (x_2 \u003d a \\). We need to find values \u200b\u200b\\ (a \\), in which the equation will have at least two roots, taking into account the fact that \\ (a\u003e 0 \\). Answer: Task 4 # 1232 Find all the parameter values \u200b\u200b\\ (a \\), each you are Task level: equal to ege It has a single decision. \
domestic and left part of the equation on \\ (2 ^ (\\ sqrt (x + 1)) \\) (because \\ (2 ^ (\\ sqrt (x + 1))\u003e 0 \\)) and rewrite the equation in the form : Consider a function \
\\ (y \u003d 2 ^ t \\ cdot \\ log _ (\\ FRAC (1) (9)) ((t + 2)) \\) When \\ (t \\ geqslant 0 \\) (because \\ (\\ SQRT (X + 1) \\ GEQSLANT 0 \\)). Derivative \\ (y "\u003d \\ left (-2 ^ t \\ cdot \\ log_9 ((t + 2)) \\ right)" \u003d - \\ dfrac (2 ^ t) (\\ ln9) \\ Cdot \\ left (\\ ln 2 \\ cdot \\ ln ((T + 2)) + \\ DFRAC (1) (T + 2) \\ Right) \\) \\ (2 ^ T\u003e 0, \\ \\ dfrac (1) (t + 2)\u003e 0, \\ \\ ln ((t + 2))\u003e 0 \\). Because With all \\ (t \\ geqslant 0 \\), then \\ (y " Therefore, when \\ (t \\ geqslant 0 \\), the function \\ (Y \\) monotonously decreases.<0\)
при всех \(t\geqslant 0\)
. The equation can be considered as \\ (y (t) \u003d y (z) \\), where \\ (z \u003d ax, t \u003d \\ sqrt (x + 1) \\). From the monotony of the function it follows that equality is possible only if \\ (t \u003d z \\). It means that the equation is equivalent to the equation: \\ (ax \u003d \\ sqrt (x + 1) \\), which in turn is equivalent to the system: \\ [\\ Begin (Cases) a ^ 2x ^ 2-X - 1 \u003d 0 \\\\ AX \\ GEQSLANT 0 \\ END (CASES) \\] When \\ (a \u003d 0 \\), the system has one solution \\ (x \u003d -1 \\), which satisfies the condition \\ (AX \\ GEQSLANT 0 \\). Consider the case \\ (A \\ NE 0 \\). Discriminant of the first equation of the system \\ (d \u003d 1 + 4a ^ 2\u003e 0 \\) with all \\ (a \\). Consequently, the equation always has two roots \\ (x_1 \\) and \\ (x_2 \\), and they are different signs (because Vieta theorem \\ (x_1 \\ cdot x_2 \u003d - \\ dfrac (1) (a ^ 2) This means that when \\ (a<0\)
). 0 \\) The condition is suitable for a positive root. Consequently, the system always has a single solution.<0\)
условию \(ax\geqslant 0\)
подходит отрицательный корень, при \(a>So, \\ (a \\ in \\ mathbb (R) \\). \\ (a \\ in \\ mathbb (R) \\). Answer: Task 5 # 1234 Task level: equal to ege It has a single decision. \
there is at least one root of the segment \\ ([- 1; 0] \\). \\ (y \u003d 2 ^ t \\ cdot \\ log _ (\\ FRAC (1) (9)) ((t + 2)) \\) \\ (f (x) \u003d 2x ^ 3-3x (AX + X-A ^ 2-1) -3a-a ^ 3 \\) With some fixed \\ (A \\). We will find its derivative: \\ (f "(x) \u003d 6x ^ 2-6Ax-6x + 3a ^ 2 + 3 \u003d 3 (x ^ 2-2ax + a ^ 2 + x ^ 2-2x + 1) \u003d 3 ((xa) ^ 2 + (x - 1) ^ 2) \\). Note that \\ (f "(x) \\ geqslant 0 \\) at all values \u200b\u200b\\ (x \\) and \\ (a \\), and equal to \\ (0 \\) only with \\ (x \u003d a \u003d 1 \\). But with \\ (a \u003d 1 \\): So, with all \\ (a \\ ne 1 \\), the function \\ (F (x) \\) is strictly increasing, therefore, the equation \\ (F (x) \u003d 0 \\) may have no more than one root. Considering the properties of the cubic function, the graph \\ (F (x) \\) with some fixed \\ (a \\) will look like this: Therefore, in order for the equation to have a root of a segment \\ ([- 1; 0] \\), it is necessary: \\ [\\ begin (Cases) f (0) \\ geqslant 0 \\\\ f (-1) \\ leqslant 0 \\ end (Cases) \\ Rightarrow \\ Begin (Cases) A \u200b\u200b(a ^ 2 + 3) \\ leqslant 0 \\\\ ( A + 2) (A ^ 2 + A + 4) \\ GEQSLANT 0 \\ END (Cases) \\ Rightarrow \\ Begin (Cases) A \u200b\u200b\\ Leqslant 0 \\\\ A \\ GEQSLANT -2 \\ END (Cases) \\ Rightarrow -2 \\ Leqslant A \\ Leqslant 0 \\] Thus, \\ (a \\ in [-2; 0] \\). Answer: \\ (A \\ in [-2; 0] \\). Task 6 # 2949 Task level: equal to ege It has a single decision. \\ [(\\ sin ^ 2x-5 \\ sin x-2a (\\ sin x-3) +6) \\ CDOT (\\ SQRT2A + 8X \\ SQRT (2x-2x ^ 2)) \u003d 0 \\] it has roots. (Task from subscribers) OST equations: \\ (2x-2x ^ 2 \\ geqslant 0 \\ quad \\ leftrightarrow \\ quad x \\ in \\). Therefore, in order for the equation to have roots, it is necessary that at least one of the equations \\ [\\ sin ^ 2x-5 \\ sin x-2a (\\ sin x-3) + 6 \u003d 0 \\ quad (\\ small (\\ text (or)) \\ quad \\ sqrt2a + 8x \\ sqrt (2x-2x ^ 2) \u003d 0 \\] dealt with OTZ. 1) Consider the first equation \\ [\\ sin ^ 2x-5 \\ Sin x-2a (\\ sin x-3) + 6 \u003d 0 \\ quad \\ leftrightarrow \\ quad \\ left [\\ begin (gathered) \\ begin (aligned) \\ sin x \u003d 2a + 2 \\\\ \\ END X \u003d 3 \\ END (aligned) \\ end (gathered) \\ RIGHT. \\ quad \\ leftrightarrow \\ quad \\ sin x \u003d 2a + 2 \\] This equation should have roots on / (\\). Consider a circle: Thus, we see that for any \\ (2a + 2 \\ in [\\ sin 0; \\ sin 1] \\) the equation will have one solution, and for all others - will not have solutions. Consequently, for \\ (a \\ in \\ left [-1; -1+ \\ sin 1 \\ right] \\) The equation has a solution. 2) Consider the second equation \\ [\\ sqrt2a + 8x \\ SQRT (2x-2x ^ 2) \u003d 0 \\ quad \\ leftrightarrow \\ quad 8x \\ sqrt (x - x ^ 2) \u003d - a \\] Consider the function \\ (f (x) \u003d 8x \\ sqrt (x - x ^ 2) \\). We will find its derivative: \
On OTZ, the derivative has one zero: \\ (X \u003d \\ FRAC34 \\), which is also the point of the maximum function \\ (F (x) \\). Therefore, in order for the equation to solve, it is necessary that the graph \\ (f (x) \\) intersects with a straight \\ (y \u003d -a \\) (one of the appropriate options is depicted in the figure). That is, you need to \
. With these \\ (x \\): The function \\ (y_1 \u003d \\ sqrt (x-1) \\) is strictly increasing. The graph of the function \\ (y_2 \u003d 5x ^ 2-9x \\) is a parabola, the vertex of which is at point \\ (X \u003d \\ DFRAC (9) (10) \\). Therefore, with all \\ (x \\ geqslant 1 \\), the function \\ (y_2 \\) also strictly increases (the right branch of the parabola). Because The sum of strictly increasing functions is strictly increasing, then \\ (F_a (x) \\) - strictly increases (constant \\ (3a + 8 \\) does not affect the monotony of the function). The function \\ (g_a (x) \u003d \\ dfrac (a ^ 2) (x) \\) with all \\ (X \\ GEQSLANT 1 \\) is part of the right branch of hyperbole and is strictly decreasing. Solve the equation \\ (F_a (x) \u003d g_a (x) \\) means to find the intersection points of the functions \\ (F \\) and \\ (G \\). From their opposite monotony it follows that the equation may have no more than one root. With \\ (x \\ geqslant 1 \\) \\ (F_A (X) \\ GEQSLANT 3A + 4, \\ \\ \\ 0 \\\\ Cup Answer: \\ (A \\ in (- \\ infty; -1] \\ CUP)
Consider the case when f / (x) ≥0. Consider two points x1, x2∈ (a, b) and apply Lagrange formula. The function f (x) satisfies all the conditions of this theorem. It follows that x1 
We now formulate without proof enough inflection condition.
It means that the equality \\ (f (t) \u003d f (u) \\) is possible if and only if \\ (t \u003d u \\). Let us return to the initial variables and solve the resulting equation:
\
\
\
Consequently, the answer: \\ (a \\ in (0; + \\ infty) \\).
\\ ((0; + \\ infty) \\).
\\ (f "(x) \u003d 6 (x-1) ^ 2 \\ rightarrow f (x) \u003d 2 (x-1) ^ 3 \\ rightarrow \\) The equation \\ (2 (x - 1) ^ 3 \u003d 0 \\) has the only root \\ (x \u003d 1 \\), which does not satisfy the condition. Consequently, \\ (A \\) cannot be equal to \\ (1 \\).
Note that \\ (f (0) \u003d f (1) \u003d 0 \\). So, schematically graph \\ (f (x) \\) looks like this: 
