Goals:
- General education: to systematize, generalize, expand the knowledge and skills of students related to the application of methods for solving inequalities.
- Developing: to develop in students the ability to listen to a lecture, concisely writing it down in a notebook.
- Educational: to form cognitive motivation to study mathematics.
During the classes
I. Introductory conversation:
We have finished the topic “Solving irrational equations” and today we are starting to learn how to solve irrational inequalities.
First, let's remember what types of inequalities you can solve and with what methods?
Answer: Linear, square, rational, trigonometric. Linear ones are solved based on the properties of inequalities, trigonometric ones are reduced to the simplest trigonometric ones, solved using a trigonometric circle, and the rest, mainly by the method of intervals.
Question: On what statement is the method of intervals based?
Answer: On a theorem stating that a continuous function that does not vanish on some interval retains its sign on that interval.
II. Let's consider an irrational inequality like >
Question: Is it possible to apply the interval method to solve it?
Answer: Yes, since the function y=- continuous on D(y).
We solve this inequality interval method .
Conclusion: we quite easily solved this irrational inequality by the interval method, actually reducing it to solving an irrational equation.
Let's try to solve another inequality with this method.
3)f(x) continuous on D(f)
4) Function zeros:
- Long search D(f).
- It is difficult to calculate breakpoints.
The question arises: “Are there other ways to solve this inequality?”.
Obviously, there is, and now we will get to know them.
III. So, topic today's lesson: "Methods for solving irrational inequalities."
The lesson will be held in the form of a lecture, since the textbook does not contain a detailed analysis of all methods. Therefore, our important task is to draw up a detailed summary of this lecture.
IV. We have already talked about the first method for solving irrational inequalities.
It - interval method , a universal method for solving all types of inequalities. But it does not always lead to the goal in a short and simple way.
v. When solving irrational inequalities, you can use the same ideas as when solving irrational equations, but since a simple verification of solutions is impossible (after all, solutions to inequalities are most often integer numerical intervals), it is necessary to use equivalence.
We present schemes for solving the main types of irrational inequalities method of equivalent transitions from one inequality to a system of inequalities.
2. It is proved similarly that
Let's write these diagrams on the reference board. Think about Type 3 and 4 proofs at home, we will discuss them in the next lesson.
VI. Let's solve the inequality in a new way.
The original inequality is equivalent to a set of systems.
VII. And there is a third method that often helps solve complex irrational inequalities. We have already talked about it in relation to inequalities with a modulus. it function substitution method (multiplier substitution). Let me remind you that the essence of the replacement method is that the difference in the values of monotonic functions can be replaced by the difference in the values of their arguments.
Consider an irrational inequality of the form<,
that is -< 0.
By theory, if p(x) increases on some interval to which they belong a and b, and a>b, then the inequalities p(a) – p(b) > 0 and a-b> 0 are equivalent to D(p), that is
VIII. We solve the inequality by the method of changing factors.
Hence, this inequality is equivalent to the system
Thus, we have seen that the use of the method of replacement of factors to reduce the solution of the inequality to the method of intervals significantly reduces the amount of work.
IX. Now that we have covered the three basic methods for solving equations, let's do independent work with self-examination.
It is necessary to perform the following numbers (according to the textbook by A. M. Mordkovich): 1790 (a) - solve_ by the method of_ equivalent transitions,_ 1791 (a) - solve by the method of replacing factors. To solve irrational inequalities, it is proposed to use the methods previously analyzed when solving irrational equations:
- change of variables;
- use of ODZ;
- use of monotonicity properties of functions.
The completion of the study of the topic is a test.
Analysis of the control work shows:
- typical errors of weak students, in addition to arithmetic and algebraic ones, are incorrect equivalent transitions to a system of inequalities;
- the method of substitution of factors is successfully used only by strong students.
Any inequality, which includes a function under the root, is called irrational. There are two types of such inequalities:
In the first case, the root is less than the function g (x), in the second - more. If g(x) - constant, the inequality simplifies dramatically. Please note that outwardly these inequalities are very similar, but their solution schemes are fundamentally different.
Today we will learn how to solve irrational inequalities of the first type - they are the simplest and most understandable. The inequality sign can be strict or non-strict. The following statement is true for them:
Theorem. Any irrational inequality of the form
Equivalent to the system of inequalities:
Not weak? Let's look at where such a system comes from:
- f (x) ≤ g 2 (x) - everything is clear here. This is the original inequality squared;
- f(x) ≥ 0 is the ODZ of the root. Let me remind you: the arithmetic square root exists only from non-negative numbers;
- g(x) ≥ 0 is the range of the root. By squaring inequality, we burn the cons. As a result, extra roots may appear. The inequality g (x) ≥ 0 cuts them off.
Many students "go in cycles" on the first inequality of the system: f (x) ≤ g 2 (x) - and completely forget the other two. The result is predictable: wrong decision, lost points.
Since irrational inequalities are a rather complicated topic, let's analyze 4 examples at once. From elementary to really complex. All tasks are taken from the entrance exams of Moscow State University. M. V. Lomonosov.
Examples of problem solving
A task. Solve the inequality:
We have a classic irrational inequality: f(x) = 2x + 3; g(x) = 2 is a constant. We have:
Only two of the three inequalities remained by the end of the solution. Because the inequality 2 ≥ 0 always holds. Let's intersect the remaining inequalities:
So, x ∈ [−1,5; 0.5]. All points are shaded because inequalities are not strict.
A task. Solve the inequality:
We apply the theorem:
We solve the first inequality. To do this, we will open the square of the difference. We have:
2x 2 − 18x + 16< (x
− 4) 2 ;
2x 2 − 18x + 16< x
2 − 8x
+ 16:
x 2 − 10x< 0;
x (x − 10)< 0;
x ∈ (0; 10).
Now let's solve the second inequality. There too square trinomial:
2x 2 − 18x + 16 ≥ 0;
x 2 − 9x + 8 ≥ 0;
(x − 8)(x − 1) ≥ 0;
x ∈ (−∞; 1]∪∪∪∪)
