The interval method is considered to be universal for solving inequalities. Sometimes this method is also called the gap method. It can be used both for solving rational inequalities with one variable and for inequalities of other types. In our material we tried to pay attention to all aspects of the issue.

What awaits you in this section? We will analyze the interval method and consider algorithms for solving inequalities using it. Let us touch upon the theoretical aspects on which the application of the method is based.

We pay special attention to the nuances of the topic that are usually not covered in the school curriculum. For example, let's consider the rules for arranging signs on intervals and the method of intervals itself in general form, without its connection to rational inequalities.

Yandex.RTB R-A-339285-1

Algorithm

Who remembers how the method of intervals was introduced in a school algebra course? Usually it all starts with solving inequalities of the form f (x)< 0 (знак неравенства может быть использован любой другой, например, ≤ , >or ≥). Here f(x) can be a polynomial or a ratio of polynomials. The polynomial, in turn, can be represented as:

• product of linear binomials with coefficient 1 for the variable x;
• the product of quadratic trinomials with leading coefficient 1 and the negative discriminant of their roots.

Here are some examples of such inequalities:

(x + 3) · (x 2 − x + 1) · (x + 2) 3 ≥ 0,

(x - 2) · (x + 5) x + 3 > 0,

(x − 5) · (x + 5) ≤ 0,

(x 2 + 2 x + 7) (x - 1) 2 (x 2 - 7) 5 (x - 1) (x - 3) 7 ≤ 0.

Let us write an algorithm for solving inequalities of this type, as we have given in the examples, using the interval method:

• we find the zeros of the numerator and denominator, for this we equate the numerator and denominator of the expression on the left side of the inequality to zero and solve the resulting equations;
• we determine the points that correspond to the found zeros and mark them with dashes on the coordinate axis;
• define expression signs f(x) from the left side of the inequality being solved on each interval and put them on the graph;
• we apply shading over the required sections of the graph, guided by the following rule: if the inequality has signs< или ≤ изображается, штрихуются «минусовые» промежутки, если же мы работаем с неравенством, имеющим знаки >or ≥ , then we highlight by shading the areas marked with the “+” sign.

The pattern we will work with may have a schematic view. Excessive details can overload the drawing and make it difficult to solve. We will be of little interest in scale. It will be enough to adhere to the correct location of the points as the values ​​of their coordinates increase.

When working with strict inequalities, we will use the notation of a point in the form of a circle with an unfilled (empty) center. In the case of non-strict inequalities, we will depict the points that correspond to the zeros of the denominator as empty, and all the rest as ordinary black.

The marked points divide the coordinate line into several numerical intervals. This allows us to obtain a geometric representation of a numerical set, which is actually a solution to this inequality.

The Science of the Gap Method

The approach underlying the interval method is based on the following property of a continuous function: the function maintains a constant sign on the interval (a, b) on which this function is continuous and does not vanish. The same property is characteristic of numerical rays (− ∞ , a) and (a, + ∞).

This property of the function is confirmed by the Bolzano-Cauchy theorem, which is given in many textbooks for preparing for entrance examinations.

The constancy of the sign on intervals can also be justified on the basis of the properties of numerical inequalities. For example, take the inequality x - 5 x + 1 > 0. If we find the zeros of the numerator and denominator and plot them on the number line, we will get a series of intervals: (− ∞ , − 1) , (− 1 , 5) and (5 , + ∞) .

Let us take any of the intervals and show on it that throughout the entire interval the expression on the left side of the inequality will have a constant sign. Let this be the interval (− ∞ , − 1) . Let's take any number t from this interval. It will satisfy the conditions t< − 1 , и так как − 1 < 5 , то по свойству транзитивности, оно же будет удовлетворять и неравенству t < 5 .

Using both the resulting inequalities and the property of numerical inequalities, we can assume that t + 1< 0 и t − 5 < 0 . Это значит, что t + 1 и t − 5 – это отрицательные числа независимо от значения t on the interval (− ∞ , − 1) .

Using the rule for dividing negative numbers, we can state that the value of the expression t - 5 t + 1 will be positive. This means that the value of the expression x - 5 x + 1 will be positive for any value x from between (− ∞ , − 1) . All this allows us to assert that on the interval taken as an example, the expression has a constant sign. In our case, this is the “+” sign.

Finding the zeros of the numerator and denominator

The algorithm for finding zeros is simple: we equate the expressions from the numerator and denominator to zero and solve the resulting equations. If you have any difficulties, you can refer to the topic “Solving equations by factorization.” In this section we will limit ourselves to just looking at an example.

Consider the fraction x · (x - 0, 6) x 7 · (x 2 + 2 · x + 7) 2 · (x + 5) 3. In order to find the zeros of the numerator and denominator, we equate them to zero in order to obtain and solve the equations: x (x − 0, 6) = 0 and x 7 (x 2 + 2 x + 7) 2 (x + 5) 3 = 0.

In the first case, we can go to the set of two equations x = 0 and x − 0, 6 = 0, which gives us two roots 0 and 0, 6. These are the zeros of the numerator.

The second equation is equivalent to the set of three equations x 7 = 0, (x 2 + 2 x + 7) 2 = 0, (x + 5) 3 = 0 . We carry out a series of transformations and get x = 0, x 2 + 2 · x + 7 = 0, x + 5 = 0. The root of the first equation is 0, the second equation has no roots, since it has a negative discriminant, the root of the third equation is 5. These are the zeros of the denominator.

0 in this case is both the zero of the numerator and the zero of the denominator.

In general, when the left side of an inequality contains a fraction that is not necessarily rational, the numerator and denominator are also equal to zero to obtain the equations. Solving the equations allows you to find the zeros of the numerator and denominator.

Determining the sign of an interval is simple. To do this, you can find the value of the expression from the left side of the inequality for any arbitrarily selected point from a given interval. The resulting sign of the expression value at an arbitrarily chosen point in the interval will coincide with the sign of the entire interval.

Let's look at this statement with an example.

Let's take the inequality x 2 - x + 4 x + 3 ≥ 0. The expression on the left side of the inequality has no zeros in the numerator. The zero of the denominator will be the number - 3. We get two intervals on the number line (− ∞ , − 3) and (− 3 , + ∞) .

In order to determine the signs of the intervals, we calculate the value of the expression x 2 - x + 4 x + 3 for points taken arbitrarily on each of the intervals.

From the first gap (− ∞ , − 3) let's take − 4. At x = − 4 we have (- 4) 2 - (- 4) + 4 (- 4) + 3 = - 24. We received a negative value, which means the entire interval will have a “-” sign.

For the gap (− 3 , + ∞) Let's carry out calculations with a point having a zero coordinate. At x = 0 we have 0 2 - 0 + 4 0 + 3 = 4 3. We received a positive value, which means that the entire interval will have a “+” sign.

You can use another way to determine signs. To do this, we can find the sign on one of the intervals and save it or change it when passing through zero. In order to do everything correctly, it is necessary to follow the rule: when passing through zero the denominator, but not the numerator, or the numerator, but not the denominator, we can change the sign to the opposite one, if the degree of the expression giving this zero is odd, and we cannot change the sign , if the degree is even. If we have received a point that is both the zero of the numerator and the denominator, then we can change the sign to the opposite one only if the sum of the powers of the expressions giving this zero is odd.

If we recall the inequality that we examined at the beginning of the first paragraph of this material, then on the rightmost interval we can put a “+” sign.

Now let's look at examples.

Take the inequality (x - 2) · (x - 3) 3 · (x - 4) 2 (x - 1) 4 · (x - 3) 5 · (x - 4) ≥ 0 and solve it using the interval method. To do this, we need to find the zeros of the numerator and denominator and mark them on the coordinate line. The zeros of the numerator will be points 2 , 3 , 4 , denominator point 1 , 3 , 4 . Let's mark them on the coordinate axis with dashes.

We mark the zeros of the denominator with empty dots.

Since we are dealing with a non-strict inequality, we replace the remaining dashes with ordinary dots.

Now let's place dots on the intervals. The rightmost space (4 , + ∞) will be a + sign.

Moving from right to left, we will put down signs for the remaining intervals. We pass through the point with coordinate 4. This is both the zero of the numerator and the denominator. In sum, these zeros give the expressions (x − 4) 2 And x − 4. Let's add their powers 2 + 1 = 3 and get an odd number. This means that the sign during the transition in this case changes to the opposite. The interval (3, 4) will have a minus sign.

We pass to the interval (2, 3) through the point with coordinate 3. This is also a zero for both the numerator and the denominator. We got it thanks to two expressions (x − 3) 3 and (x − 3) 5, the sum of the powers of which is 3 + 5 = 8. Getting an even number allows us to leave the sign of the interval unchanged.

The point with coordinate 2 is the zero of the numerator. The power of the expression x - 2 is 1 (odd). This means that when passing through this point the sign must be changed to the opposite.

We have the last interval left (− ∞ , 1) . The point with coordinate 1 is the zero of the denominator. It was derived from the expression (x − 1) 4, with even degree 4 . Therefore, the sign remains the same. The final drawing will look like this:

The interval method is especially effective when calculating the value of an expression involves a lot of work. An example would be the need to calculate the value of an expression

x + 3 - 3 4 3 x 2 + 6 x + 11 2 x + 2 - 3 4 (x - 1) 2 x - 2 3 5 (x - 12)

at any point in the interval 3 - 3 4, 3 - 2 4.

Now let's start applying the acquired knowledge and skills in practice.

Example 1

Solve the inequality (x - 1) · (x + 5) 2 (x - 7) · (x - 1) 3 ≤ 0.

Solution

It is advisable to use the interval method to solve the inequality. Find the zeros of the numerator and denominator. The zeros of the numerator are 1 and - 5, the zeros of the denominator are 7 and 1. Let's mark them on the number line. We are dealing with a non-strict inequality, so we will mark the zeros of the denominator with empty dots, and the zero of the numerator - 5 - will be marked with a regular filled dot.

Let's put the signs of the intervals using the rules for changing the sign when passing through zero. Let's start with the rightmost interval, for which we calculate the value of the expression from the left side of the inequality at a point arbitrarily taken from the interval. We get the “+” sign. Let's move sequentially through all the points on the coordinate line, arranging the signs, and get:

We work with a non-strict inequality with the sign ≤. This means that we need to mark with shading the spaces marked with the “-” sign.

Answer: (- ∞ , 1) ∪ (1 , 7) .

Solving rational inequalities in most cases requires their preliminary transformation to the desired form. Only after this it becomes possible to use the interval method. Algorithms for carrying out such transformations are discussed in the material “Solving rational inequalities.”

Let's look at an example of converting quadratic trinomials into inequalities.

Example 2

Find the solution to the inequality (x 2 + 3 x + 3) (x + 3) x 2 + 2 x - 8 > 0.

Solution

Let's see if the discriminants of the quadratic trinomials in the inequality notation are actually negative. This will allow us to determine whether the form of this inequality allows us to use the interval method for solution.

Let's calculate the discriminant for the trinomial x 2 + 3 x + 3: D = 3 2 − 4 1 3 = − 3< 0 . Now let’s calculate the discriminant for the trinomial x 2 + 2 · x − 8: D ’ = 1 2 − 1 · (− 8) = 9 > 0 . As you can see, the inequality requires a preliminary transformation. To do this, we represent the trinomial x 2 + 2 x − 8 as (x + 4) · (x − 2), and then apply the interval method to solve the inequality (x 2 + 3 · x + 3) · (x + 3) (x + 4) · (x - 2) > 0.

Answer: (- 4 , - 3) ∪ (2 , + ∞) .

The generalized interval method is used to solve inequalities of the form f (x)< 0 (≤ , >, ≥) , where f (x) is an arbitrary expression with one variable x.

All actions are carried out according to a certain algorithm. In this case, the algorithm for solving inequalities using the generalized interval method will be slightly different from what we discussed earlier:

• we find the domain of definition of the function f and the zeros of this function;
• mark the boundary points on the coordinate axis;
• plot the zeros of the function on the number line;
• determine the signs of intervals;
• apply shading;
• write down the answer.

On the number line, it is necessary to mark, among other things, individual points of the domain of definition. For example, the domain of definition of a function is the set (− 5, 1 ] ∪ ( 3 ) ∪ [ 4 , 7) ∪ ( 10 ) . This means that we need to mark points with coordinates − 5, 1, 3, 4 , 7 And 10 . Points − 5 and 7 will be depicted as empty, the rest can be highlighted with a colored pencil in order to distinguish them from the zeros of the function.

In the case of non-strict inequalities, the zeros of the function are plotted by ordinary (shaded) points, and in the case of strict inequalities, by empty points. If the zeros coincide with the boundary points or individual points of the domain of definition, then they can be repainted black, making them empty or shaded, depending on the type of inequality.

The response record is a numerical set that includes:

• spaces with shading;
• individual points of the domain of definition with a plus sign, if we are dealing with an inequality whose sign is > or ≥, or with a minus sign, if the inequality has signs< или ≤ .

Now it has become clear that the algorithm that we presented at the very beginning of the topic is a special case of the algorithm for using the generalized interval method.

Let's consider an example of using the generalized interval method.

Example 3

Solve the inequality x 2 + 2 x - 24 - 3 4 x - 3 x - 7< 0 .

Solution

We introduce a function f such that f (x) = x 2 + 2 x - 24 - 3 4 x - 3 x - 7 . Let's find the domain of definition of the function f:

x 2 + 2 x - 24 ≥ 0 x ≠ 7 D (f) = (- ∞ , - 6 ] ∪ [ 4 , 7) ∪ (7 , + ∞) .

Now let's find the zeros of the function. To do this, we will solve the irrational equation:

x 2 + 2 x - 24 - 3 4 x - 3 = 0

We get the root x = 12.

To indicate boundary points on the coordinate axis, we use orange. Points - 6, 4 will be filled in, and 7 will be left empty. We get:

Let's mark the zero of the function with an empty black dot, since we are working with a strict inequality.

We determine the signs at individual intervals. To do this, take one point from each interval, for example, 16 , 8 , 6 And − 8 , and calculate the value of the function in them f:

f (16) = 16 2 + 2 16 - 24 - 3 4 16 - 3 16 - 7 = 264 - 15 9 > 0 f (8) = 8 2 + 2 8 - 24 - 3 4 8 - 3 8 - 7 = 56 - 9< 0 f (6) = 6 2 + 2 · 6 - 24 - 3 4 · 6 - 3 6 - 7 = 24 - 15 2 - 1 = = 15 - 2 · 24 2 = 225 - 96 2 >0 f (- 8) = - 8 2 + 2 · (- 8) - 24 - 3 4 · (- 8) - 3 - 8 - 7 = 24 + 3 - 15< 0

We place the newly defined signs and apply shading over the spaces with a minus sign:

The answer will be the union of two intervals with the “-” sign: (− ∞, − 6 ] ∪ (7, 12).

In response, we included a point with coordinate - 6. This is not the zero of the function, which we would not include in the answer when solving a strict inequality, but the boundary point of the domain of definition, which is included in the domain of definition. The value of the function at this point is negative, which means that it satisfies the inequality.

We did not include point 4 in the answer, just as we did not include the entire interval [4, 7). At this point, just like throughout the entire indicated interval, the value of the function is positive, which does not satisfy the inequality being solved.

Let's write this down again for a clearer understanding: colored dots must be included in the answer in the following cases:

• these points are part of the hatched gap,
• these points are individual points in the domain of definition of the function, the values ​​of the function at which satisfy the inequality being solved.

Answer: (− ∞ , − 6 ] ∪ (7 , 12) .

If you notice an error in the text, please highlight it and press Ctrl+Enter

First level

Interval method. The Ultimate Guide (2019)

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, it 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 to one side; after all, we only need to leave zero on the right 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 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:

1. We move everything to the left side, leaving only zero on the right;
2. We find ODZ;
3. We plot all the roots of the inequality on the axis;
4. 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;
5. 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. AVERAGE 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:

That's what it is 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 out 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:

1. 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!

You can use our tasks (optional) and we, of course, recommend them.

In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.

How? There are two options:

1. Unlock all hidden tasks in this article - 299 rub.
2. Unlock access to all hidden tasks in all 99 articles of the textbook - 999 rub.

Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.

In the second case we will give you simulator “6000 problems with solutions and answers, for each topic, at all levels of complexity.” It will definitely be enough to get your hands on solving problems on any topic.

In fact, this is much more than just a simulator - a whole training program. If necessary, you can also use it for FREE.

Access to all texts and programs is provided for the ENTIRE period of the site's existence.

In conclusion...

If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

Interval method is a special algorithm designed to solve complex inequalities of the form f(x) > 0. The algorithm consists of 5 steps:

1. Solve the equation f(x) = 0. Thus, instead of an inequality, we get an equation that is much simpler to solve;
2. Mark all obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find the multiplicity of the roots. If the roots are of even multiplicity, then draw a loop above the root. (A root is considered a multiple if there are an even number of identical solutions)
4. Find out the sign (plus or minus) of the function f(x) on the rightmost interval. To do this, it is enough to substitute into f(x) any number that will be to the right of all the marked roots;
5. Mark the signs at the remaining intervals, alternating them.

After this, all that remains is to write down the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f(x) > 0, or with a “−” sign if the inequality was of the form f(x)< 0.

In the case of non-strict inequalities (≤ , ≥), it is necessary to include in the intervals points that are a solution to the equation f(x) = 0;

Example 1:

Solve inequality:

(x - 2)(x + 7)< 0

We work using the interval method.

Step 1: replace the inequality with an equation and solve it:

(x - 2)(x + 7) = 0

The product is zero if and only if at least one of the factors is zero:

x - 2 = 0 => x = 2

x + 7 = 0 => x = -7

We got two roots.

Step 2: We mark these roots on the coordinate line. We have:

Step 3: we find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that is greater than the number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10 and even x = 10,000).

f(x) = (x - 2)(x + 7)

f(3)=(3 - 2)(3 + 7) = 1*10 = 10

We get that f(3) = 10 > 0 (10 is a positive number), so we put a plus sign in the rightmost interval.

Step 4: you need to note the signs on the remaining intervals. We remember that when passing through each root the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus to the left. This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, to the left of the root x = −7 there is a plus. It remains to mark these signs on the coordinate axis.

Let's return to the original inequality, which had the form:

(x - 2)(x + 7)< 0

So the function must be less than zero. This means that we are interested in the minus sign, which appears only on one interval: (−7; 2). This will be the answer.

Example 2:

Solve inequality:

(9x 2 - 6x + 1)(x - 2) ≥ 0

Solution:

First you need to find the roots of the equation

(9x 2 - 6x + 1)(x - 2) = 0

Let's collapse the first bracket and get:

(3x - 1) 2 (x - 2) = 0

x - 2 = 0; (3x - 1) 2 = 0

Solving these equations we get:

Let's plot the points on the number line:

Because x 2 and x 3 are multiple roots, then there will be one point on the line and above it “ a loop”.

Let's take any number less than the leftmost point and substitute it into the original inequality. Let's take the number -1.

Don’t forget to include the solution to the equation (found X), because our inequality is not strict.

Answer: () U ∪[-6;4]∪\left\(6\right\)\)

Example.(Assignment from the OGE) Solve the inequality using the interval method \(x^2 (-x^2-64)≤64(-x^2-64)\)
Solution:

Answer : \((-∞;-8]∪}