Galaeva Ekaterina, student of the 11th grade of the MAOU secondary school No. 149, Nizhny Novgorod

The work is both applied and research in nature. For the sake of completeness, the following questions were considered:

– How are the properties of a function reflected when solving equations and inequalities?

– What equations and inequalities are solved through the definition of the properties of the domain of definition, the set of values, invariance?

– What is the solution algorithm?

- The tasks with the parameter proposed in the KIM materials in preparation for the exam were considered.

In her work, Ekaterina explored a wide range of tasks and systematized them according to their appearance.

Download:

Preview:

https://accounts.google.com

Slides captions:

Solve the inequality Solution. The function f (x) = monotonically increases on the entire real line, and the function g (x) = monotonically decreases throughout the entire domain of definition. Therefore, the inequality f (x) > g (x) is satisfied if x >

Thank you for your attention!

Preview:

To use the preview of presentations, create a Google account (account) and sign in: https://accounts.google.com

Slides captions:

Application of function properties when solving equations and inequalities Completed the work: Galaeva Ekaterina MBOU secondary school No. 149 of the Moskovsky district Pupils of 11 "A" class Supervisor: Fadeeva I. A. Mathematics teacher

Main directions: Studying the properties of a function: monotonicity, boundedness, domain of definition and invariance Learn the main statements that are most often used in solving equations, inequalities and systems Solving problems from KIM materials to prepare for the exam

Monotonicity A function increases if a larger value of the argument corresponds to a larger value of the function. The function is decreasing if the larger value of the argument corresponds to the smaller value of the function. f(x 1) f(x 2) x 1 x 2 f(x 1) f(x 2) x 1 x 2

Statement 1. If the function y \u003d f (x) is monotone, then the equation f (x) \u003d c has at most one root. x =2 f(x) = - monotonically decreasing, so there are no other solutions. Answer: x=2

Statement 2. If the function y \u003d f (x) is monotonically increasing, and the function y \u003d g (x) is monotonically decreasing, then the equation f (x) \u003d g (x) has at most one root. 2 - x \u003d lg (x + 11) + 1 g (x) \u003d 2 - x is monotonically decreasing, and the function f (x) \u003d log (x + 11) + 1 is monotonically increasing on the domain, which means that the equation f (x ) = g (x) has at most one root. By selection, we determine that x \u003d -1. The above assertion substantiates the uniqueness of the solution.

a) f (x) ≤ g (x) if and only if x ϵ (- ∞ ; x 0 ]; b) f (x) ≥ g (x) if and only if x ϵ [x 0; +∞). The visual meaning of this statement is obvious. Statement 3. If the function y \u003d f (x) is monotonically increasing on the entire real line, the function y \u003d g (x) is monotonically decreasing on the entire real line and f (x 0) \u003d g (x 0), then the following statements are true:

Solve the inequality Solution. The function f (x) = monotonically increases on the entire real line, and the function g (x) = monotonically decreases throughout the entire domain of definition. Therefore, the inequality f (x) > g (x) is satisfied if x > 2. Let's add the domain of the inequality. Thus, we get the system Answer: (2; 5).

Statement 4. If the function y \u003d f (x) is monotonically increasing, then the equations f (x) \u003d x and f (f (x)) \u003d x have the same set of roots, regardless of the number of investments. Consequence. If n is a natural number, and the function y \u003d f (x) is monotonically increasing, then the equations f (x) \u003d x and n times have the same set of roots.

Solve the equation. Answer: Decision. For x ≥1, the right side of the equation is not less than 1, and the left side is less than 1. Therefore, if the equation has roots, then any of them is less than 1. For x ≤0, the right side of the equation is non-positive, and the left side is positive, due to the fact that . Thus, any root of this equation belongs to the interval (0; 1) Multiplying both sides of this equation by x, and dividing the numerator and denominator of the left side by x, we get

Where = . Denoting through t, where t 0, we get the equation = t. Consider a function f (t)= 1+ increasing on its domain of definition. The resulting equation can be written as f (f (f (f (t))))= t , and by the corollary of Statement 4, it has the same set of solutions as the equation f (t)= t , i.e. equation 1 + = t, whence. The only positive root of this quadratic equation is . So, where, i.e. , or. Answer:

Statement 1. If max f (x) = c and min g (x) = c, then the equation f (x) = g (x) has the same set of solutions as the system Boundedness The maximum value of the left side is 1 and the minimum value the right side 1 , which means that the solution of the equation is reduced to the system of equations: , from the second equation we find a possible candidate x=0 , and we make sure that it is a solution to the first equation. Answer: x=1 .

Solve the equation Solution. Since sin3x≤1 and cos4x≤1, the left side of this equation does not exceed 7. It can be equal to 7 if and only if whence where k , n ϵ Z . It remains to establish whether there exist such integers k and n for which the latter system has solutions. Answer: Z

In problems with unknown x and parameter a, the domain of definition is understood as the set of all ordered pairs of numbers (x ; a), each of which is such that after substituting the corresponding values ​​of x and a into all relations included in the problem, they will be determined. Example 1. For each value of the parameter a, solve the inequality Solution. Let us find the domain of definition of this inequality. From which it is clear that the system has no solutions. This means that the domain of definition of the inequality does not contain any pairs of numbers x and a, and therefore the inequality has no solutions. Scope Answer:

Invariance, i.e. the invariance of an equation or inequality with respect to the replacement of a variable by some algebraic expression of this variable. The simplest example of invariance is parity: if is an even function, then the equation is invariant under the change of x and – x, since = 0. Invariance

Find the roots of the equation. Solution. Note that the pair is invariant under replacement. Replacing in equality, we get. Multiplying both sides of this equality by 2 and subtracting the equality term by term from the resulting equality, we find 3, whence. Now it remains to solve the equation, from where the roots of the equation are numbers. Answer: .

Find all values ​​of a for each of which the equation has more than three different solutions. Solving problems with the Monotonicity property parameter

|x|= positive X= |x|= For two roots to exist, the numerator must be positive. Therefore, When the roots of the first and second equations coincide, which does not meet the requirement of the condition: the presence of more than three roots. Answer: .

Find all values ​​of a for each of which the equation has two roots. Let's transform the equation to the form AND consider the function f(x)= defined and continuous on the entire real line. The graph of this function is a broken line, consisting of line segments and rays, each link of which is part of a straight line of the form y= kt+l . f(x)= For any expansion of the module of the first expression, k does not exceed 8, so the increase and decrease of the function f(x) will depend on the expansion of the second module. At x, f(x) will decrease, and at x, it will increase. That is, at x=3 the function will take the largest value. In order for the equation to have two roots, it is necessary that f(3) Monotonicity property

f(3)=12- |9-| 3+a || | 9-| 3+a || 9- | 3+a | - | 3+a | | 3+a | | 3+a | 3+a a Answer: a

Find all values ​​of the parameter a, for each of which the inequality is satisfied for any real value x Let us rewrite the inequality in the form, introduce a new variable t = and consider the function f (t) = , defined and continuous on the entire real line. The graph of this function is a broken line, consisting of line segments and rays, each link of which is part of a straight line, where to

Since, then t ϵ [-1; one]. Due to the monotonic decrease of the function y = f (t), it suffices to check the left edge of this segment. Z. A is true Means that it is possible only if the numbers u and v have the same sign or any of them is equal to zero. , = () () 0. Factoring the square trinomials, we obtain the inequality (, from which we find that a ϵ (-∞; -1] U (2) U [ 4; +∞). Answer: (-∞; -1]U(2)U)