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You just need to understand this method and know it like the back of your hand! If only because it is used to solve rational inequalities and because, knowing this method properly, solving these inequalities is surprisingly simple. A little later I’ll tell you a couple of secrets on how to save time solving these inequalities. Well, are you intrigued? Then let's go!
The essence of the method is to factor the inequality into factors (repeat the topic) and determine the ODZ and the sign of the factors; now I’ll explain everything. Let's take the simplest example: .
There is no need to write the range of acceptable values () here, since there is no division by the variable, and there are no radicals (roots) observed here. Everything here is already factorized for us. But don’t relax, this is all to remind you of the basics and understand the essence!
Let's say you don't know the interval method, how would you solve this inequality? Approach logically and build on what you already know. Firstly, the left side will be greater than zero if both expressions in parentheses are either greater than zero or less than zero, because “plus” for “plus” gives “plus” and “minus” for “minus” gives “plus”, right? And if the signs of the expressions in brackets are different, then in the end the left side will be less than zero. What do we need to find out those values at which the expressions in brackets will be negative or positive?
We need to solve an equation, it is exactly the same as an inequality, only instead of a sign there will be a sign, the roots of this equation will allow us to determine those boundary values, when departing from which the factors will be greater or less than zero.
And now the intervals themselves. What is an interval? This is a certain interval of the number line, that is, all possible numbers contained between two numbers - the ends of the interval. It’s not so easy to imagine these intervals in your head, so it’s common to draw intervals, I’ll teach you now.
We draw an axis; the entire number series from and to is located on it. Points are plotted on the axis, the very so-called zeros of the function, the values at which the expression equals zero. These points are "pinned out" which means that they are not among those values at which the inequality is true. In this case, they are punctured because sign in the inequality and not, that is, strictly greater than and not greater than or equal to.
I want to say that it is not necessary to mark zero, it is here without circles, but just for understanding and orientation along the axis. Okay, we’ve drawn the axis, put the dots (more precisely, circles), what next, how will this help me in solving? - you ask. Now just take the value for x from the intervals in order and substitute them into your inequality and see what sign the multiplication results in.
In short, we just take it for example, substitute it here, it will work out, which means that the inequality will be valid over the entire interval (over the entire interval) from to, from which we took it. In other words, if x is from to, then the inequality is true.
We do the same with the interval from to, take or, for example, substitute in, determine the sign, the sign will be “minus”. And we do the same with the last, third interval from to, where the sign turns out to be “plus”. There’s such a lot of text, but not enough clarity, right?
Take another look at inequality.
Now we also apply the signs that will be obtained as a result on the same axis. In my example, a broken line denotes the positive and negative sections of the axis.
Look at the inequality - at the drawing, again at the inequality - and again at the drawing, is anything clear? Now try to say on what intervals X, the inequality will be true. That's right, from to the inequality will also be true from to, but on the interval from to the inequality is zero and this interval is of little interest to us, because we have a sign in the inequality.
Well, now that you’ve figured it out, the only thing left to do is write down the answer! In response, we write those intervals for which the left side is greater than zero, which reads as X belongs to the interval from minus infinity to minus one and from two to plus infinity. It is worth clarifying that the parentheses mean that the values by which the interval is limited are not solutions to the inequality, that is, they are not included in the answer, but only indicate that up to, for example, is not a solution.
Now an example in which you will not only have to draw the interval:
What do you think needs to be done before putting points on the axis? Yeah, factor it into factors:
We draw intervals and place signs, notice that we have punctured dots because the sign is strictly less than zero:
It's time to tell you one secret that I promised at the beginning of this topic! What if I told you that you don’t have to substitute the values from each interval to determine the sign, but you can determine the sign in one of the intervals, and simply alternate the signs in the rest!
Thus, we saved a little time on putting down signs - I think this gained time on the Unified State Exam will not hurt!
We write the answer:
Now consider an example of a fractional-rational inequality - an inequality, both parts of which are rational expressions (see).
What can you say about this inequality? And you look at it as a fractional-rational equation, what do we do first? We immediately see that there are no roots, which means it’s definitely rational, but then it’s a fraction, and even with an unknown in the denominator!
That's right, we need ODZ!
So, let's go further, here all the factors except one have a variable of the first degree, but there is a factor where x has a second degree. Usually, our sign changed after passing through one of the points at which the left side of the inequality takes on a zero value, for which we determined what x should be equal to in each factor. But here, it’s always positive, because any number squared > zero and a positive term.
Do you think this will affect the meaning of inequality? That's right - it won't affect! We can safely divide the inequality into both parts and thereby remove this factor so that it is not an eyesore.
The time has come to draw the intervals; to do this, you need to determine those boundary values, when departing from which the multipliers will be greater and less than zero. But pay attention that there is a sign here, which means that we will not pick out the point at which the left side of the inequality takes on a zero value, it is included in the number of solutions, we have only one such point, this is the point where x is equal to one. Shall we color the point where the denominator is negative? - Of course not!
The denominator must not be zero, so the interval will look like this:
Using this diagram, you can easily write the answer, I’ll just say that now you have a new type of bracket at your disposal - square! Here's a bracket [ says that the value is included in the solution interval, i.e. is part of the answer, this bracket corresponds to a filled (not pinned) point on the axis.
So, did you get the same answer?
We factor it into factors and move everything in one direction; after all, we only need to leave zero on the right in order to compare with it:
I draw your attention to the fact that in the last transformation, in order to obtain in the numerator as well as in the denominator, I multiply both sides of the inequality by. Remember that when both sides of an inequality are multiplied by, the sign of the inequality changes to the opposite!!!
We write ODZ:
Otherwise, the denominator will go to zero, and, as you remember, you cannot divide by zero!
Agree, the resulting inequality is tempting to reduce the numerator and denominator! This cannot be done; you may lose some of the decisions or ODZ!
Now try to put the points on the axis yourself. I will only note that when plotting points, you need to pay attention to the fact that a point with a value, which, based on the sign, would seem to be plotted on the axis as shaded, will not be shaded, it will be gouged out! Why do you ask? And remember the ODZ, you’re not going to divide by zero like that?
Remember, ODZ comes first! If all the inequalities and equal signs say one thing, and the ODZ says another, trust the ODZ, great and powerful! Well, you built the intervals, I'm sure you took my hint about alternation and you got it like this (see picture below) Now cross it out and don't make that mistake again! What error? - you ask.
The fact is that in this inequality the factor was repeated twice (remember how you tried to reduce it?). So, if some factor is repeated in the inequality an even number of times, then when passing through a point on the axis that turns this factor to zero (in this case, a point), the sign will not change; if it is odd, then the sign changes!
The following axis with intervals and signs will be correct:
And, please note that the sign we are interested in is not the one that was at the beginning (when we first saw the inequality, the sign was there), after the transformations, the sign changed to, which means we are interested in intervals with a sign.
Answer:
I will also say that there are situations when there are roots of inequality that are not included in any interval, in response they are written in curly brackets, like this, for example: . You can read more about such situations in the article average level.
Let's summarize how to solve inequalities using the interval method:
- We move everything to the left side, leaving only zero on the right;
- We find ODZ;
- We plot all the roots of the inequality on the axis;
- We take an arbitrary one from one of the intervals and determine the sign in the interval to which the root belongs, alternate the signs, paying attention to the roots that are repeated several times in the inequality; whether the sign changes when passing through them depends on the evenness or oddness of the number of times they are repeated or not;
- In response, we write intervals, observing the punctured and non-punctured points (see ODZ), placing the necessary types of brackets between them.
And finally, our favorite section, “do it yourself”!
Examples:
Answers:
INTERVAL METHOD. MIDDLE LEVEL
Linear function
A function of the form is called linear. Let's take a function as an example. It is positive at and negative at. The point is the zero of the function (). Let's show the signs of this function on the number axis:
We say that “the function changes sign when passing through the point”.
It can be seen that the signs of the function correspond to the position of the function graph: if the graph is above the axis, the sign is “ ”, if below it is “ ”.
If we generalize the resulting rule to an arbitrary linear function, we obtain the following algorithm:
- Finding the zero of the function;
- We mark it on the number axis;
- We determine the sign of the function on opposite sides of zero.
Quadratic function
I hope you remember how to solve quadratic inequalities? If not, read the topic. Let me remind you of the general form of a quadratic function: .
Now let's remember what signs the quadratic function takes. Its graph is a parabola, and the function takes the sign " " for those in which the parabola is above the axis, and " " - if the parabola is below the axis:
If a function has zeros (values at which), the parabola intersects the axis at two points - the roots of the corresponding quadratic equation. Thus, the axis is divided into three intervals, and the signs of the function alternately change when passing through each root.
Is it possible to somehow determine the signs without drawing a parabola every time?
Recall that a square trinomial can be factorized:
For example: .
Let's mark the roots on the axis:
We remember that the sign of a function can only change when passing through the root. Let's use this fact: for each of the three intervals into which the axis is divided by roots, it is enough to determine the sign of the function at only one arbitrarily chosen point: at the remaining points of the interval the sign will be the same.
In our example: at both expressions in brackets are positive (substitute, for example:). We put a “ ” sign on the axis:
Well, when (substitute, for example), both brackets are negative, which means the product is positive:
This is it interval method: knowing the signs of the factors on each interval, we determine the sign of the entire product.
Let's also consider cases when the function has no zeros, or only one.
If they are not there, then there are no roots. This means that there will be no “passing through the root”. This means that the function takes only one sign on the entire number line. It can be easily determined by substituting it into a function.
If there is only one root, the parabola touches the axis, so the sign of the function does not change when passing through the root. What rule can we come up with for such situations?
If you factor such a function, you get two identical factors:
And any squared expression is non-negative! Therefore, the sign of the function does not change. In such cases, we will highlight the root, when passing through which the sign does not change, by circling it with a square:
We will call such a root a multiple.
Interval method in inequalities
Now any quadratic 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 intervals depending on the sign of the inequality. For example:
Let's measure the roots on the axis and place the signs:
We need the part of the axis with the " " sign; since the inequality is not strict, the roots themselves are also included in the solution:
Now consider a rational inequality - an inequality, both sides of which are rational expressions (see).
Example:
All factors except one are “linear” here, that is, they contain a variable only to the first power. We need such linear factors to apply the interval method - the sign changes when passing through their roots. But the multiplier has no roots at all. This means that it is always positive (check this for yourself), and therefore does not affect the sign of the entire inequality. This means that we can divide the left and right sides of the inequality by it, and thus get rid of it:
Now everything is the same as it was with quadratic inequalities: we determine at what points each of the factors becomes zero, mark these points on the axis and arrange the signs. I would like to draw your attention to a very important fact:
Answer: . Example: .
To apply the interval method, one of the parts of the inequality must have. Therefore, let's move the right side to the left:
The numerator and denominator have the same factor, but don’t rush to reduce it! After all, then we may forget to prick out this point. It is better to mark this root as a multiple, that is, when passing through it, the sign will not change: 
Answer: .
And one more very illustrative example:
Again, we don't cancel the same factors of the numerator and denominator, because if we do, we'll have to specifically remember to puncture the dot.
- : repeated times;
- : times;
- : times (in the numerator and one in the denominator).
In the case of an even number, we do the same as before: we draw a square around the point and do not change the sign when passing through the root. But in the case of an odd number, this rule does not apply: the sign will still change when passing through the root. Therefore, we do not do anything additional with such a root, as if it were not a multiple. The above rules apply to all even and odd powers. 
What should we write in the answer?
If the alternation of signs is violated, you need to be very careful, because if the inequality is not strict, the answer should include all shaded points. But some of them often stand apart, that is, they are not included in the shaded area. In this case, we add them to the answer as isolated points (in curly braces):
Examples (decide for yourself):
Answers:
- If among the factors it is simple, it is a root, because it can be represented as.
.
INTERVAL METHOD. BRIEFLY ABOUT THE MAIN THINGS
The interval method is used to solve rational inequalities. It consists in determining the sign of the product from the signs of the factors on various intervals.
Algorithm for solving rational inequalities using the interval method.
- We move everything to the left side, leaving only zero on the right;
- We find ODZ;
- We plot all the roots of the inequality on the axis;
- We take an arbitrary one from one of the intervals and determine the sign in the interval to which the root belongs, alternate the signs, paying attention to the roots that are repeated several times in the inequality; whether the sign changes when passing through them depends on the evenness or oddness of the number of times they are repeated or not;
- In response, we write intervals, observing the punctured and non-punctured points (see ODZ), placing the necessary types of brackets between them.
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
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It has been necessary to compare quantities and quantities when solving practical problems since ancient times. At the same time, words such as more and less, higher and lower, lighter and heavier, quieter and louder, cheaper and more expensive, etc. appeared, denoting the results of comparing homogeneous quantities.
The concepts of more and less arose in connection with counting objects, measuring and comparing quantities. For example, mathematicians of Ancient Greece knew that the side of any triangle is less than the sum of the other two sides and that the larger side lies opposite the larger angle in a triangle. Archimedes, while calculating the circumference, established that the perimeter of any circle is equal to three times the diameter with an excess that is less than a seventh of the diameter, but more than ten seventy times the diameter.
Symbolically write relationships between numbers and quantities using the signs > and b. Records in which two numbers are connected by one of the signs: > (greater than), You also encountered numerical inequalities in the lower grades. You know that inequalities can be true, or they can be false. For example, \(\frac(1)(2) > \frac(1)(3)\) is a correct numerical inequality, 0.23 > 0.235 is an incorrect numerical inequality.
Inequalities involving unknowns may be true for some values of the unknowns and false for others. For example, the inequality 2x+1>5 is true for x = 3, but false for x = -3. For an inequality with one unknown, you can set the task: solve the inequality. In practice, problems of solving inequalities are posed and solved no less often than problems of solving equations. For example, many economic problems come down to the study and solution of systems of linear inequalities. In many branches of mathematics, inequalities are more common than equations.
Some inequalities serve as the only auxiliary means of proving or disproving the existence of a certain object, for example, the root of an equation.
Numerical inequalities
You can compare whole numbers and decimal fractions. Know the rules for comparing ordinary fractions with the same denominators but different numerators; with the same numerators but different denominators. Here you will learn how to compare any two numbers by finding the sign of their difference.
Comparing numbers is widely used in practice. For example, an economist compares planned indicators with actual ones, a doctor compares a patient’s temperature with normal, a turner compares the dimensions of a machined part with a standard. In all such cases, some numbers are compared. As a result of comparing numbers, numerical inequalities arise.
Definition. The number a is greater than the number b if the difference a-b is positive. The number a is less than the number b if the difference a-b is negative.
If a is greater than b, then they write: a > b; if a is less than b, then they write: a Thus, the inequality a > b means that the difference a - b is positive, i.e. a - b > 0. Inequality a For any two numbers a and b from the following three relations a > b, a = b, a To compare the numbers a and b means to find out which of the signs >, = or Theorem. If a > b and b > c, then a > c.
Theorem. If you add the same number to both sides of the inequality, the sign of the inequality will not change.
Consequence. Any term can be transferred from one part of the inequality to another by changing the sign of this term to the opposite.
Theorem. If both sides of the inequality are multiplied by the same positive number, then the sign of the inequality will not change. If both sides of the inequality are multiplied by the same negative number, then the sign of the inequality will change to the opposite.
Consequence. If both sides of the inequality are divided by the same positive number, then the sign of the inequality will not change. If both sides of the inequality are divided by the same negative number, then the sign of the inequality will change to the opposite.
You know that numerical equalities can be added and multiplied term by term. Next, you will learn how to perform similar actions with inequalities. The ability to add and multiply inequalities term by term is often used in practice. These actions help solve problems of evaluating and comparing the meanings of expressions.
When solving various problems, it is often necessary to add or multiply the left and right sides of inequalities term by term. At the same time, it is sometimes said that inequalities add up or multiply. For example, if a tourist walked more than 20 km on the first day, and more than 25 km on the second, then we can say that in two days he walked more than 45 km. Similarly, if the length of a rectangle is less than 13 cm and the width is less than 5 cm, then we can say that the area of this rectangle is less than 65 cm2.
When considering these examples, the following were used: theorems on addition and multiplication of inequalities:
Theorem. When adding inequalities of the same sign, an inequality of the same sign is obtained: if a > b and c > d, then a + c > b + d.
Theorem. When multiplying inequalities of the same sign, whose left and right sides are positive, an inequality of the same sign is obtained: if a > b, c > d and a, b, c, d are positive numbers, then ac > bd.
Inequalities with the sign > (greater than) and 1/2, 3/4 b, c Along with the signs of strict inequalities > and In the same way, the inequality \(a \geq b \) means that the number a is greater than or equal to b, i.e. .and not less b.
Inequalities containing the \(\geq \) sign or the \(\leq \) sign are called non-strict. For example, \(18 \geq 12 , \; 11 \leq 12 \) are not strict inequalities.
All properties of strict inequalities are also valid for non-strict inequalities. Moreover, if for strict inequalities the signs > were considered opposite and you know that to solve a number of applied problems you have to create a mathematical model in the form of an equation or a system of equations. Next, you will learn that mathematical models for solving many problems are inequalities with unknowns. The concept of solving an inequality will be introduced and how to test whether a given number is a solution to a particular inequality will be shown.
Inequalities of the form
\(ax > b, \quad ax in which a and b are given numbers, and x is an unknown, are called linear inequalities with one unknown.
Definition. The solution to an inequality with one unknown is the value of the unknown at which this inequality becomes a true numerical inequality. Solving an inequality means finding all its solutions or establishing that there are none.
You solved the equations by reducing them to the simplest equations. Similarly, when solving inequalities, one tries to reduce them, using properties, to the form of simple inequalities.
Solving second degree inequalities with one variable
Inequalities of the form
\(ax^2+bx+c >0 \) and \(ax^2+bx+c where x is a variable, a, b and c are some numbers and \(a \neq 0 \), called inequalities of the second degree with one variable.
Solution to inequality
\(ax^2+bx+c >0 \) or \(ax^2+bx+c can be considered as finding intervals in which the function \(y= ax^2+bx+c \) takes positive or negative values To do this, it is enough to analyze how the graph of the function \(y= ax^2+bx+c\) is located in the coordinate plane: where the branches of the parabola are directed - up or down, whether the parabola intersects the x axis and if it does, then at what points.
Algorithm for solving second degree inequalities with one variable:
1) find the discriminant of the square trinomial \(ax^2+bx+c\) and find out whether the trinomial has roots;
2) if the trinomial has roots, then mark them on the x-axis and through the marked points draw a schematic parabola, the branches of which are directed upward for a > 0 or downward for a 0 or at the bottom for a 3) find intervals on the x-axis for which the points parabolas are located above the x-axis (if they solve the inequality \(ax^2+bx+c >0\)) or below the x-axis (if they solve the inequality
\(ax^2+bx+c Solving inequalities using the interval method
Consider the function
f(x) = (x + 2)(x - 3)(x - 5)
The domain of this function is the set of all numbers. The zeros of the function are the numbers -2, 3, 5. They divide the domain of definition of the function into the intervals \((-\infty; -2), \; (-2; 3), \; (3; 5) \) and \( (5; +\infty)\)
Let us find out what the signs of this function are in each of the indicated intervals.
The expression (x + 2)(x - 3)(x - 5) is the product of three factors. The sign of each of these factors in the intervals under consideration is indicated in the table:
In general, let the function be given by the formula
f(x) = (x-x 1)(x-x 2) ... (x-x n),
where x is a variable, and x 1, x 2, ..., x n are numbers that are not equal to each other. The numbers x 1 , x 2 , ..., x n are the zeros of the function. In each of the intervals into which the domain of definition is divided by zeros of the function, the sign of the function is preserved, and when passing through zero its sign changes.
This property is used to solve inequalities of the form
(x-x 1)(x-x 2) ... (x-x n) > 0,
(x-x 1)(x-x 2) ... (x-x n) where x 1, x 2, ..., x n are numbers not equal to each other
Considered method solving inequalities is called the interval method.
Let us give examples of solving inequalities using the interval method.
Solve inequality:
\(x(0.5-x)(x+4) Obviously, the zeros of the function f(x) = x(0.5-x)(x+4) are the points \(x=0, \; x= \frac(1)(2) , \; x=-4 \)We plot the zeros of the function on the number axis and calculate the sign on each interval:
We select those intervals at which the function is less than or equal to zero and write down the answer.
Answer:
\(x \in \left(-\infty; \; 1 \right) \cup \left[ 4; \; +\infty \right) \)
In this lesson we will continue solving rational inequalities using the interval method for more complex inequalities. Let us consider the solution of fractional linear and fractional quadratic inequalities and related problems.
Now let's return to the inequality
Let's look at some related tasks.
Find the smallest solution to the inequality.
Find the number of natural solutions to the inequality
Find the length of the intervals that make up the set of solutions to the inequality.
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1. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of general education institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 28(b,c); 29(b,c); 35(a,b); 37(b,c); 38(a).
Interval method(or as it is sometimes called the interval method) is a universal method for solving inequalities. It is suitable for solving a variety of inequalities, but is most convenient in solving rational inequalities with one variable. Therefore, in the school algebra course, the method of intervals is closely tied specifically to rational inequalities, and practically no attention is paid to solving other inequalities with its help.
In this article we will analyze the interval method in detail and touch on all the intricacies of solving inequalities with one variable using it. Let's start by presenting an algorithm for solving inequalities using the interval method. Next, we will explain what theoretical aspects it is based on and analyze the steps of the algorithm, in particular, we will dwell in detail on the determination of signs on intervals. After this, we will move on to practice and show solutions to several typical examples. And in conclusion, we will consider the interval method in general form (that is, without reference to rational inequalities), in other words, the generalized interval method.
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Algorithm
Acquaintance with the interval method in school begins with solving inequalities of the form f(x)<0
(знак неравенства может быть и другим ≤, >or ≥), where f(x) is either , represented as a product linear binomials with 1 for variable x and/or square trinomials with a leading coefficient of 1 and with a negative discriminant and their degrees, or the ratio of such polynomials. For clarity, we give examples of such inequalities: (x−5)·(x+5)≤0 , (x+3)·(x 2 −x+1)·(x+2) 3 ≥0, 
.
To make further conversation substantive, let’s immediately write down an algorithm for solving inequalities of the above type using the interval method, and then we’ll figure out what, how and why. So, using the interval method:
- First, the zeros of the numerator and zeros of the denominator are found. To do this, the numerator and denominator of the expression on the left side of the inequality are equal to zero, and the resulting equations are solved.
- After this, the points corresponding to the found zeros are marked with dashes. A schematic drawing is enough, in which it is not necessary to observe the scale, the main thing is to adhere to the location of the points relative to each other: the point with the smaller coordinate is located to the left of the point with the larger coordinate. After this, it becomes clear how they should be depicted: regular or punctured (with an empty center). When solving a strict inequality (with sign< или >) all points are depicted as punctured. When solving a non-strict inequality (with a sign ≤ or ≥), the points corresponding to the zeros of the denominator are punctured, and the remaining points marked with dashes are ordinary. These points divide the coordinate line into several numerical intervals.
- Next, the signs of the expression f(x) are determined from the left side of the inequality being solved on each interval (we will describe in detail how this is done in one of the following paragraphs), and + or − are placed above them in accordance with the signs defined on them.
- Finally, when solving the signed inequality< или ≤ изображается штриховка над промежутками, отмеченными знаком −, а при решении неравенства со знаком >or ≥ - over spaces marked with a + sign. The result is , which is the desired solution to the inequality.
Note that the above algorithm is consistent with the description of the interval method in school textbooks.
What is the method based on?
The approach underlying the interval method takes place due to the following property of a continuous function: if on the interval (a, b) the function f is continuous and does not vanish, then it retains a constant sign on this interval (we would add that a similar property this is also true for the number rays (−∞, a) and (a, +∞) ). And this property, in turn, follows from the Bolzano-Cauchy theorem (its consideration is beyond the scope of the school curriculum), the formulation and proof of which, if necessary, can be found, for example, in the book.
For expressions f(x) having the form indicated in the previous paragraph, the constancy of the sign on intervals can be justified in another way, starting from the properties of numerical inequalities and taking into account the rules for multiplying and dividing numbers with the same signs and different signs.
As an example, consider the inequality. The zeros of its numerator and denominator divide the number line into three intervals (−∞, −1), (−1, 5) and (5, +∞). Let us show that on the interval (−∞, −1) the expression on the left side of the inequality has a constant sign (we can take another interval, the reasoning will be similar). Let's take any number t from this interval. It will obviously satisfy the inequality t<−1 , и так как −1<5 , то по свойству транзитивности, оно же будет удовлетворять и неравенству t<5 . Из этих неравенств в силу свойств числовых неравенств следует, что t+1<0 и t−5<0. То есть, t+1 и t−5 – отрицательные числа, не зависимо от того, какое конкретно число t мы возьмем из промежутка (−∞, −1) . Тогда позволяет констатировать, что значение выражения будет положительным, откуда следует, что значение выражения будет положительным при любом значении x из промежутка (−∞, −1) . Итак, на указанном промежутке выражение имеет постоянный знак, причем, это знак +.
So we smoothly approached the issue of determining signs on intervals, but we will not skip over the first step of the interval method, which involves finding the zeros of the numerator and denominator.
How to find the zeros of the numerator and denominator?
Finding the zeros of the numerator and denominator of a fraction of the type indicated in the first paragraph usually does not pose any problems. For this, the expressions from the numerator and denominator are set equal to zero, and the resulting equations are solved. The principle of solving equations of this type is described in detail in the article solving equations by factorization method. Here we will just limit ourselves to an example.
Consider the fraction 
and find the zeros of its numerator and denominator. Let's start with the zeros of the numerator. We equate the numerator to zero, we obtain the equation x·(x−0.6)=0, from which we proceed to the set of two equations x=0 and x−0.6=0, from where we find two roots 0 and 0.6. These are the required zeros of the numerator. Now we find the zeros of the denominator. Let's make an equation x 7 ·(x 2 +2·x+7) 2 ·(x+5) 3 =0, it is equivalent to a set of three equations x 7 =0, (x 2 +2 x+7) 2 =0, (x+5) 3 =0, and then x=0, x 2 +2 x+7=0 , x+5=0 . The root of the first of these equations is obvious, it is 0, the second equation has no roots, since its discriminant is negative, and the root of the third equation is −5. So, we found the zeros of the denominator, there were two of them: 0 and −5. Note that 0 turned out to be both a zero in the numerator and a zero in the denominator.
To find the zeros of the numerator and denominator in the general case, when the left side of the inequality is a fraction, but not necessarily rational, the numerator and denominator are also equated to zero, and the corresponding equations are solved.
How to determine signs at intervals?
The most reliable way to determine the sign of the expression on the left side of the inequality on each interval is to calculate the value of this expression at any one point in each interval. In this case, the desired sign on the interval coincides with the sign of the value of the expression at any point in this interval. Let's explain this with an example.
Let's take inequality 
. The expression on its left side has no zeros in the numerator, and the zero in the denominator is the number −3. It divides the number line into two intervals (−∞, −3) and (−3, +∞). Let's determine the signs on them. To do this, take one point from these intervals and calculate the values of the expression in them. Let us immediately note that it is advisable to take such points so that it is easy to carry out calculations. For example, from the first interval (−∞, −3) we can take −4. For x=−4 we have 
, received a value with a minus sign (negative), therefore, there will be a minus sign on this interval. We move on to determining the sign on the second interval (−3, +∞). It is convenient to take 0 from it (if 0 is included in the interval, then it is advisable to always take it, since at x=0 the calculations are the simplest). At x=0 we have 
. This value has a plus sign (positive), so there will be a plus sign on this interval.
There is another approach to determining signs, which consists of finding the sign at one of the intervals and maintaining it or changing it when moving to the adjacent interval through zero. You must adhere to the following rule. When passing through the zero of the numerator, but not the denominator, or through the zero of the denominator, but not the numerator, the sign changes if the degree of the expression giving this zero is odd, and does not change if it is even. And when passing through a point that is both the zero of the numerator and the zero of the denominator, the sign changes if the sum of the powers of the expressions giving this zero is odd, and does not change if it is even.
By the way, if the expression on the right side of the inequality has the form indicated at the beginning of the first paragraph of this article, then there will be a plus sign on the rightmost gap.
To make everything clear, let's look at an example.
Let there be inequality before us 
, and we solve it using the interval method. To do this, we find the zeros of the numerator 2, 3, 4 and the zeros of the denominator 1, 3, 4, mark them on the coordinate line first with dashes 
then we replace the zeros of the denominator with images of punctured dots 
and since we are solving a non-strict inequality, we replace the remaining dashes with ordinary dots 
And then comes the moment of identifying signs at intervals. As we noticed before this example, on the rightmost interval (4, +∞) there will be a + sign: 
Let's determine the remaining signs, while moving from gap to gap from right to left. Moving on to the next interval (3, 4), we pass through the point with coordinate 4. This is the zero of both the numerator and the denominator, these zeros give the expressions (x−4) 2 and x−4, the sum of their powers is 2+1=3, and this is an odd number, which means that when passing through this point you need to change the sign. Therefore, on the interval (3, 4) there will be a minus sign: 
We go further to the interval (2, 3), while passing through the point with coordinate 3. This is also the zero of both the numerator and the denominator, it is given by the expressions (x−3) 3 and (x−3) 5, the sum of their powers is 3+5=8, and this is an even number, therefore, the sign will remain unchanged: 
We move further to the interval (1, 2). The path to it is blocked by a point with coordinate 2. This is the zero of the numerator, it is given by the expression x−2, its degree is 1, that is, it is odd, therefore, when passing through this point, the sign will change: 
Finally, it remains to determine the sign on the last interval (−∞, 1) . To get to it, we need to overcome the point with coordinate 1. This is the zero of the denominator, it is given by the expression (x−1) 4, its degree is 4, that is, it is even, therefore, the sign will not change when passing through this point. So we have identified all the signs, and the drawing takes on the following form: 
It is clear that the use of the considered method is especially justified when calculating the value of an expression involves a large amount of work. For example, calculate the value of the expression 
at any point in the interval 
.
Examples of solving inequalities using the interval method
Now you can put together all the information presented, sufficient to solve inequalities using the interval method, and analyze the solutions of several examples.
Example.
Solve the inequality 
.
Solution.
Let us solve this inequality using the interval method. Obviously, the zeros of the numerator are 1 and −5, and the zeros of the denominator are 1. We mark them on the number line, with the points with coordinates and 1 pricked out as zeros of the denominator, and the remaining zero of the numerator −5 is represented by an ordinary point, since we are solving a non-strict inequality: 
Now we put signs on the intervals, adhering to the rule of maintaining or changing the sign when passing through zeros. There will be a + sign above the rightmost gap (this can be checked by calculating the value of the expression on the left side of the inequality at some point in this gap, for example, at x=3). When passing through the sign we change, when passing through 1 we leave it the same, and when passing through −5 we again leave the sign unchanged: 
Since we are solving the inequality with the ≤ sign, it remains to draw shading over the intervals marked with the sign − and write down the answer from the resulting image. 
So, the solution we are looking for is: 
.
Answer:
.
To be fair, let us draw your attention to the fact that in the overwhelming majority of cases, when solving rational inequalities, they first have to be transformed to the required form in order to make it possible to solve them using the method of intervals. We will discuss in detail how to carry out such transformations in the article. solving rational inequalities, and now we will give an example illustrating one important point regarding quadratic trinomials in the recording of inequalities.
Example.
Find the solution to the inequality 
.
Solution.
At first glance, this inequality appears to be of a form suitable for applying the interval method. But it doesn’t hurt to check whether the discriminants of the quadratic trinomials in his notation are really negative. Let's figure them out to ease our conscience. For the trinomial x 2 +3 x+3 we have D=3 2 −4 1 3=−3<0
, а для трехчлена x 2 +2·x−8
получаем D’=1 2 −1·(−8)=9>0 . This means that transformations are required to give this inequality the desired form. In this case, it is enough to represent the trinomial x 2 +2 x−8 as (x+4) (x−2) , and then solve the inequality using the method of intervals 
.
Answer:
.
Generalized interval method
The generalized interval method allows you to solve inequalities of the form f(x)<0 (≤, >, ≥), where f(x) is arbitrary with one variable x. Let's write it down algorithm for solving inequalities using the generalized interval method:
- First you need f and the zeros of this function.
- The boundary points, including individual points, of the domain of definition are marked on the number line. For example, if the domain of a function is the set (−5, 1]∪(3)∪ (we do not define the sign on the interval (−6, 4), since it is not part of the domain of definition of the function). To do this, take one point from each interval, for example, 16, 8 , 6 and −8, and calculate the value of the function f in them:
If you have questions about how it was found out what the calculated values of the function are, positive or negative, then study the material in the article comparison of numbers.
We place the just defined signs, and apply shading over the spaces with a minus sign:
In the answer we write the union of two intervals with the sign −, we have (−∞, −6]∪(7, 12). Note that −6 is included in the answer (the corresponding point is solid, not punctured). The fact is that this not the zero of the function (which, when solving a strict inequality, we would not include in the answer), but the boundary point of the domain of definition (it is colored, not black), and the value of the function at this point is negative (as evidenced by the minus sign). over the corresponding interval), that is, it satisfies the inequality. But 4 does not need to be included in the answer (as well as the entire interval ∪(7, 12) .
References.
- Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
- Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
- Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
- Kudryavtsev L. D. Course of mathematical analysis (in two volumes): Textbook for university and college students. – M.: Higher. school, 1981, vol. 1. – 687 p., ill.
Interval method– a simple way to solve fractional rational inequalities. This is the name for inequalities containing rational (or fractional-rational) expressions that depend on a variable.
1. Consider, for example, the following inequality
The interval method allows you to solve it in a couple of minutes.
On the left side of this inequality is a fractional rational function. Rational because it does not contain roots, sines, or logarithms - only rational expressions. On the right is zero.
The interval method is based on the following property of a fractional rational function.
A fractional rational function can change sign only at those points at which it is equal to zero or does not exist.
Let us recall how a quadratic trinomial is factorized, that is, an expression of the form .
Where and are the roots of the quadratic equation.
We draw an axis and place the points at which the numerator and denominator go to zero.
The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded, since the inequality is not strict. When and our inequality is satisfied, since both its sides are equal to zero.
These points break the axis into intervals.
Let us determine the sign of the fractional rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points at which it is equal to zero or does not exist. This means that at each of the intervals between the points where the numerator or denominator goes to zero, the sign of the expression on the left side of the inequality will be constant - either “plus” or “minus”.
And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that is convenient for us.
. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.
Next interval: . Let's check the sign at . We find that the left side has changed its sign to .
Let's take it. When the expression is positive - therefore, it is positive over the entire interval from to.
When the left side of the inequality is negative.
And finally, class="tex" alt="x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}
We have found at what intervals the expression is positive. All that remains is to write down the answer:
Answer: .
Please note: the signs alternate between intervals. This happened because when passing through each point, exactly one of the linear factors changed sign, while the rest kept it unchanged.
We see that the interval method is very simple. To solve the fractional-rational inequality using the interval method, we reduce it to the form:
Or class="tex" alt="\genfrac())()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or , or .
(on the left side is a fractional rational function, on the right side is zero).
Then we mark on the number line the points at which the numerator or denominator goes to zero.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
All that remains is to find out its sign at each interval.
We do this by checking the sign of the expression at any point belonging to a given interval. After that, we write down the answer. That's it.But the question arises: do the signs always alternate? No, not always! You must be careful and not place signs mechanically and thoughtlessly.
2. Let's consider another inequality.
Class="tex" alt="\genfrac())()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3 \right))>0"> !}
Place the points on the axis again. The dots and are punctured because they are zeros of the denominator. The point is also cut out, since the inequality is strict.
When the numerator is positive, both factors in the denominator are negative. This can be easily checked by taking any number from a given interval, for example, . The left side has the sign:
When the numerator is positive; The first factor in the denominator is positive, the second factor is negative. The left side has the sign:
The situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:
Finally, with class="tex" alt="x>3"> все множители положительны, и левая часть имеет знак :!}
Answer: .
Why was the alternation of signs disrupted? Because when passing through a point the multiplier is “responsible” for it didn't change sign. Consequently, the entire left side of our inequality did not change sign.
Conclusion: if the linear multiplier is an even power (for example, squared), then when passing through a point the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.
3. Let's consider a more complex case. It differs from the previous one in that the inequality is not strict:
The left side is the same as in the previous problem. The picture of signs will be the same:
Maybe the answer will be the same? No! A solution is added This happens because at both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.
Answer: .
This situation often occurs in problems on the Unified State Examination in mathematics. This is where applicants fall into a trap and lose points. Be careful!
4. What to do if the numerator or denominator cannot be factored into linear factors? Consider this inequality:
A square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression for all is the same, and specifically, positive. You can read more about this in the article on properties of quadratic functions.
And now we can divide both sides of our inequality by a value that is positive for all. Let us arrive at an equivalent inequality:
Which is easily solved using the interval method.
Please note that we divided both sides of the inequality by a value that we knew for sure was positive. Of course, in general, you should not multiply or divide an inequality by a variable whose sign is unknown.
5 . Let's consider another inequality, seemingly quite simple:
I just want to multiply it by . But we are already smart, and we won’t do this. After all, it can be both positive and negative. And we know that if both sides of the inequality are multiplied by a negative value, the sign of the inequality changes.
We will do it differently - we will collect everything in one part and bring it to a common denominator. The right side will remain zero:
Class="tex" alt="\genfrac())()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}
And after that - apply interval method.
