Spacing Method is a simple way to solve fractional rational inequalities. This is the name of 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 contains neither roots, nor sines, nor 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 where it is equal to zero or does not exist.

Recall how a square trinomial is factored, that is, an expression of the form .

Where and are the roots of the quadratic equation.

We draw an axis and arrange the points at which the numerator and denominator vanish.

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 because the inequality is not strict. For and our inequality is satisfied, since both its parts 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 where it is equal to zero or does not exist. This means that on each of the intervals between the points where the numerator or denominator vanishes, 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 suits 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 for . We get that the left side has changed sign to .

Let's take . When the expression is positive - therefore, it is positive on the entire interval from to .

For , the left side of the inequality is negative.

And finally class="tex" alt="(!LANG:x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}

We have found on what intervals the expression is positive. It remains to write the answer:

Answer: .

Please note: the signs on the intervals alternate. This happened because when passing through each point, exactly one of the linear factors changed sign, and the rest kept it unchanged.

We see that the interval method is very simple. To solve a fractional-rational inequality by the method of intervals, we bring it to the form:

Or class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or or .

(on the left side - a fractional-rational function, on the right side - zero).

Then - we mark on the number line the points at which the numerator or denominator vanishes.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
It remains only to find out its sign on each interval.
We do this by checking the sign of the expression at any point within the given interval. After that, we write down the answer. That's all.

But the question arises: do the signs always alternate? No not always! We must be careful not to place signs mechanically and thoughtlessly.

2. Let's look at another inequality.

Class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3\right))>0"> !}

We again place points on the axis. The points and are punctured because they are the zeros of the denominator. The dot is also punctured, since the inequality is strict.

When the numerator is positive, both factors in the denominator are negative. This is easy to check 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:

When 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="(!LANG:x>3"> все множители положительны, и левая часть имеет знак :!}

Answer: .

Why was the alternation of characters broken? Because when passing through the point, the multiplier "responsible" for it did not change sign. Consequently, the entire left-hand side of our inequality did not change sign either.

Conclusion: if the linear factor is in an even power (for example, in a square), 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 complicated 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? Not! The solution is added This is because at both the left and right parts of the inequality are equal to zero - therefore, this point is a solution.

Answer: .

In the problem on the exam in mathematics, this situation is often encountered. Here, applicants fall into a trap and lose points. Be careful!

4. What if the numerator or denominator cannot be factored into linear factors? Consider this inequality:

The 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 is the same for all, and specifically, it is positive. You can read more about this in the article on the properties of a quadratic function.

And now we can divide both sides of our inequality by a value that is positive for all . We arrive at an equivalent inequality:

Which is easily solved by the interval method.

Pay attention - we divided both sides of the inequality by the value, which we knew for sure that it was positive. Of course, in the general case, you should not multiply or divide an inequality by a variable whose sign is unknown.

5 . Consider another inequality, seemingly quite simple:

So I want to multiply it by . But we are already smart, and we will not do this. After all, it can be both positive and negative. And we know that if both parts of the inequality are multiplied by a negative value, the sign of the inequality changes.

We will act differently - we will collect everything in one part and bring it to a common denominator. Zero will remain on the right side:

Class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}

And after that - applicable interval method.

And today not everyone can solve rational inequalities. More precisely, not only everyone can decide. Few people can do it.
Klitschko

This lesson is going to be tough. So tough that only the Chosen will reach the end of it. Therefore, before reading, I recommend removing women, cats, pregnant children and ...

Okay, it's actually quite simple. Suppose you have mastered the interval method (if you have not mastered it, I recommend you go back and read it) and learned how to solve inequalities of the form $P\left(x \right) \gt 0$, where $P\left(x \right)$ is some polynomial or product of polynomials.

I believe that it will not be difficult for you to solve, for example, such a game (by the way, try it for a warm-up):

\[\begin(align) & \left(2((x)^(2))+3x+4 \right)\left(4x+25 \right) \gt 0; \\ & x\left(2((x)^(2))-3x-20 \right)\left(x-1 \right)\ge 0; \\ & \left(8x-((x)^(4)) \right)((\left(x-5 \right))^(6))\le 0. \\ \end(align)\]

Now let's complicate the task a little and consider not just polynomials, but the so-called rational fractions of the form:

where $P\left(x \right)$ and $Q\left(x \right)$ are the same polynomials of the form $((a)_(n))((x)^(n))+(( a)_(n-1))((x)^(n-1))+...+((a)_(0))$, or the product of such polynomials.

This will be a rational inequality. The fundamental point is the presence of the variable $x$ in the denominator. For example, here are rational inequalities:

\[\begin(align) & \frac(x-3)(x+7) \lt 0; \\ & \frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0; \\ & \frac(3((x)^(2))+10x+3)(((\left(3-x \right))^(2))\left(4-((x)^( 2)) \right))\ge 0. \\ \end(align)\]

And this is not a rational, but the most common inequality, which is solved by the interval method:

\[\frac(((x)^(2))+6x+9)(5)\ge 0\]

Looking ahead, I’ll say right away: there are at least two ways to solve rational inequalities, but all of them in one way or another are reduced to the method of intervals already known to us. Therefore, before analyzing these methods, let's recall the old facts, otherwise there will be no sense from the new material.

What you already need to know

There are not many important facts. We really only need four.

Abbreviated multiplication formulas

Yes, yes: they will haunt us throughout the school math curriculum. And at the university too. There are quite a few of these formulas, but we only need the following:

\[\begin(align) & ((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2)); \\ & ((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right); \\ & ((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b) ^(2))\right); \\ & ((a)^(3))-((b)^(3))=\left(ab \right)\left(((a)^(2))+ab+((b)^( 2))\right). \\ \end(align)\]

Pay attention to the last two formulas - this is the sum and difference of cubes (and not the cube of the sum or difference!). They are easy to remember if you notice that the sign in the first bracket is the same as the sign in the original expression, and in the second bracket it is the opposite of the sign in the original expression.

Linear equations

These are the simplest equations of the form $ax+b=0$, where $a$ and $b$ are ordinary numbers, and $a\ne 0$. This equation is easy to solve:

\[\begin(align) & ax+b=0; \\ &ax=-b; \\ & x=-\frac(b)(a). \\ \end(align)\]

I note that we have the right to divide by the coefficient $a$, because $a\ne 0$. This requirement is quite logical, since with $a=0$ we get this:

First, there is no $x$ variable in this equation. This, generally speaking, should not confuse us (this happens, say, in geometry, and quite often), but still we are no longer a linear equation.

Secondly, the solution of this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation is $0=0$. This equality is always true; hence $x$ is any number (usually written as $x\in \mathbb(R)$). If the coefficient $b$ is not equal to zero, then the equality $b=0$ is never satisfied, i.e. no answers (written $x\in \varnothing $ and read "solution set is empty").

To avoid all these complexities, we simply assume $a\ne 0$, which does not in any way restrict us from further reflections.

Quadratic equations

Let me remind you that this is called a quadratic equation:

Here on the left is a polynomial of the second degree, and again $a\ne 0$ (otherwise, instead of a quadratic equation, we get a linear one). The following equations are solved through the discriminant:

1. If $D \gt 0$, we get two different roots;
2. If $D=0$, then the root will be one, but of the second multiplicity (what kind of multiplicity it is and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
3. For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. This, by the way, is a very useful fact, which for some reason is forgotten to be told in algebra classes.

The roots themselves are calculated according to the well-known formula:

\[((x)_(1,2))=\frac(-b\pm \sqrt(D))(2a)\]

Hence, by the way, the restrictions on the discriminant. After all, the square root of a negative number does not exist. As for the roots, many students have a terrible mess in their heads, so I specially recorded a whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)

Operations with rational fractions

Everything that was written above, you already know if you studied the method of intervals. But what we will analyze now has no analogues in the past - this is a completely new fact.

Definition. A rational fraction is an expression of the form

\[\frac(P\left(x \right))(Q\left(x \right))\]

where $P\left(x \right)$ and $Q\left(x \right)$ are polynomials.

It is obvious that it is easy to obtain an inequality from such a fraction - it is enough just to attribute the sign “greater than” or “less than” to the right. And a little further we will find that solving such problems is a pleasure, everything is very simple there.

Problems begin when there are several such fractions in one expression. They have to be reduced to a common denominator - and it is at this moment that a large number of offensive mistakes are made.

Therefore, in order to successfully solve rational equations, it is necessary to firmly master two skills:

1. Factorization of the polynomial $P\left(x \right)$;
2. Actually, bringing fractions to a common denominator.

How to factorize a polynomial? Very simple. Let we have a polynomial of the form

Let's equate it to zero. We get the $n$-th degree equation:

\[((a)_(n))((x)^(n))+((a)_(n-1))((x)^(n-1))+...+(( a)_(1))x+((a)_(0))=0\]

Let's say we solved this equation and got the roots $((x)_(1)),\ ...,\ ((x)_(n))$ (don't worry: in most cases there will be no more than two of these roots) . In this case, our original polynomial can be rewritten like this:

\[\begin(align) & P\left(x \right)=((a)_(n))((x)^(n))+((a)_(n-1))((x )^(n-1))+...+((a)_(1))x+((a)_(0))= \\ & =((a)_(n))\left(x -((x)_(1)) \right)\cdot \left(x-((x)_(2)) \right)\cdot ...\cdot \left(x-((x)_( n)) \right) \end(align)\]

That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate factor in front of the brackets, and if necessary, it can be inserted into any of these brackets (practice shows that with $((a)_ (n))\ne \pm 1$ there are almost always fractions among the roots).

Task. Simplify the expression:

\[\frac(((x)^(2))+x-20)(x-4)-\frac(2((x)^(2))-5x+3)(2x-3)-\ frac(4-8x-5((x)^(2)))(x+2)\]

Solution. First, let's look at the denominators: they are all linear binomials, and there is nothing to factorize here. So let's factorize the numerators:

\[\begin(align) & ((x)^(2))+x-20=\left(x+5 \right)\left(x-4 \right); \\ & 2((x)^(2))-5x+3=2\left(x-\frac(3)(2) \right)\left(x-1 \right)=\left(2x- 3\right)\left(x-1\right); \\ & 4-8x-5((x)^(2))=-5\left(x+2 \right)\left(x-\frac(2)(5) \right)=\left(x +2 \right)\left(2-5x \right). \\\end(align)\]

Please note: in the second polynomial, the senior coefficient "2", in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since a fraction got out there.

The same thing happened in the third polynomial, only there the order of the terms is also confused. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter a factor in one and only one bracket!), which saved us from the inconvenience associated with fractional roots.

As for the first polynomial, everything is simple there: its roots are sought either in the standard way through the discriminant, or using the Vieta theorem.

Let's go back to the original expression and rewrite it with the numerators decomposed into factors:

\[\begin(matrix) \frac(\left(x+5 \right)\left(x-4 \right))(x-4)-\frac(\left(2x-3 \right)\left( x-1 \right))(2x-3)-\frac(\left(x+2 \right)\left(2-5x \right))(x+2)= \\ =\left(x+5 \right)-\left(x-1 \right)-\left(2-5x \right)= \\ =x+5-x+1-2+5x= \\ =5x+4. \\ \end(matrix)\]

Answer: $5x+4$.

As you can see, nothing complicated. A bit of 7th-8th grade math and that's it. The point of all transformations is to turn a complex and scary expression into something simple and easy to work with.

However, this will not always be the case. So now we will consider a more serious problem.

But first, let's figure out how to bring two fractions to a common denominator. The algorithm is extremely simple:

1. Factorize both denominators;
2. Consider the first denominator and add to it the factors present in the second denominator, but not in the first. The resulting product will be the common denominator;
3. Find out what factors each of the original fractions lacks so that the denominators become equal to the common one.

Perhaps this algorithm will seem to you just a text in which there are “a lot of letters”. So let's take a look at a specific example.

Task. Simplify the expression:

\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]

Solution. Such voluminous tasks are best solved in parts. Let's write out what is in the first bracket:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3))-8 )-\frac(1)(x-2)\]

Unlike the previous problem, here the denominators are not so simple. Let's factorize each of them.

The square trinomial $((x)^(2))+2x+4$ cannot be factorized because the equation $((x)^(2))+2x+4=0$ has no roots (the discriminant is negative). We leave it unchanged.

The second denominator, the cubic polynomial $((x)^(3))-8$, upon closer examination is the difference of cubes and can be easily decomposed using the abbreviated multiplication formulas:

\[((x)^(3))-8=((x)^(3))-((2)^(3))=\left(x-2 \right)\left(((x) ^(2))+2x+4 \right)\]

Nothing else can be factored, since the first bracket contains a linear binomial, and the second one is a construction already familiar to us, which has no real roots.

Finally, the third denominator is a linear binomial that cannot be decomposed. Thus, our equation will take the form:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))-\frac(1)(x-2)\]

It is quite obvious that $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$ will be the common denominator, and to reduce all fractions to it, you need to multiply the first fraction to $\left(x-2 \right)$, and the last one to $\left(((x)^(2))+2x+4 \right)$. Then it remains only to bring the following:

\[\begin(matrix) \frac(x\cdot \left(x-2 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \ right))+\frac(((x)^(2))+8)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))- \frac(1\cdot \left(((x)^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x +4 \right))= \\ =\frac(x\cdot \left(x-2 \right)+\left(((x)^(2))+8 \right)-\left(((x )^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \\ =\frac (((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))= \\ =\frac(((x)^(2))-4x+4)(\left(x-2 \right)\ left(((x)^(2))+2x+4 \right)). \\ \end(matrix)\]

Pay attention to the second line: when the denominator is already common, i.e. instead of three separate fractions, we wrote one large one, you should not immediately get rid of the brackets. It is better to write an extra line and note that, say, there was a minus before the third fraction - and it will not go anywhere, but will “hang” in the numerator in front of the bracket. This will save you a lot of mistakes.

Well, in the last line it is useful to factorize the numerator. Moreover, this is an exact square, and the abbreviated multiplication formulas again come to our aid. We have:

\[\frac(((x)^(2))-4x+4)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right) )=\frac(x-2)(((x)^(2))+2x+4)\]

Now let's deal with the second bracket in the same way. Here I will simply write a chain of equalities:

\[\begin(matrix) \frac(((x)^(2)))(((x)^(2))-4)-\frac(2)(2-x)=\frac((( x)^(2)))(\left(x-2 \right)\left(x+2 \right))-\frac(2)(2-x)= \\ =\frac(((x) ^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)= \\ =\frac(((x)^( 2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2\cdot \left(x+2 \right))(\left(x-2 \right )\cdot \left(x+2 \right))= \\ =\frac(((x)^(2))+2\cdot \left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right) ). \\ \end(matrix)\]

We return to the original problem and look at the product:

\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]

Answer: \[\frac(1)(x+2)\].

The meaning of this problem is the same as the previous one: to show how much rational expressions can be simplified if you approach their transformation wisely.

And now, when you know all this, let's move on to the main topic of today's lesson - solving fractional rational inequalities. Moreover, after such preparation, the inequalities themselves will click like nuts. :)

The main way to solve rational inequalities

There are at least two approaches to solving rational inequalities. Now we will consider one of them - the one that is generally accepted in the school mathematics course.

But first, let's note an important detail. All inequalities are divided into two types:

1. Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
2. Nonstrict: $f\left(x \right)\ge 0$ or $f\left(x \right)\le 0$.

Inequalities of the second type are easily reduced to the first, as well as the equation:

This small "addition" $f\left(x \right)=0$ leads to such an unpleasant thing as filled points - we met them back in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's analyze the universal algorithm:

1. Collect all non-zero elements on one side of the inequality sign. For example, on the left;
2. Bring all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factorize into the numerator and denominator. One way or another, we get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the tick is the inequality sign.
3. Equate the numerator to zero: $P\left(x \right)=0$. We solve this equation and get the roots $((x)_(1))$, $((x)_(2))$, $((x)_(3))$, ... Then we require that the denominator was not equal to zero: $Q\left(x \right)\ne 0$. Of course, in essence, we have to solve the equation $Q\left(x \right)=0$, and we get the roots $x_(1)^(*)$, $x_(2)^(*)$, $x_(3 )^(*)$, ... (in real problems there will hardly be more than three such roots).
4. We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punched out.
5. We place the plus and minus signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a "plus". If $f\left(x \right) \lt 0$, then we look at intervals with "minuses".

Practice shows that points 2 and 4 cause the greatest difficulties - competent transformations and the correct arrangement of numbers in ascending order. Well, at the last step, be extremely careful: we always place signs based on the last inequality written before moving on to the equations. This is a universal rule inherited from the interval method.

So, there is a scheme. Let's practice.

Task. Solve the inequality:

\[\frac(x-3)(x+7) \lt 0\]

Solution. We have a strict inequality of the form $f\left(x \right) \lt 0$. Obviously, points 1 and 2 from our scheme have already been completed: all the elements of inequality are collected on the left, nothing needs to be reduced to a common denominator. So let's move on to the third point.

Set the numerator to zero:

\[\begin(align) & x-3=0; \\ &x=3. \end(align)\]

And the denominator:

\[\begin(align) & x+7=0; \\ & ((x)^(*))=-7. \\ \end(align)\]

In this place, many people get stuck, because in theory you need to write down $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But after all, in the future we will poke out the points that came from the denominator, so you should not complicate your calculations once again - write an equal sign everywhere and don’t worry. No one will deduct points for this. :)

Fourth point. We mark the obtained roots on the number line:

All points are punctured because the inequality is strict

Note: all points are punctured because the original inequality is strict. And here it doesn’t matter anymore: these points came from the numerator or from the denominator.

Well, look at the signs. Take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but you could have just as well taken $((x)_(0))=3.1$ or $((x)_(0)) =1\000\000$). We get:

So, to the right of all the roots we have a positive area. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, we proceed to the fifth point: we place the signs and choose the right one:

We return to the last inequality, which was before solving the equations. Actually, it coincides with the original one, because we did not perform any transformations in this task.

Since it is necessary to solve an inequality of the form $f\left(x \right) \lt 0$, I shaded the interval $x\in \left(-7;3 \right)$ - it is the only one marked with a minus sign. This is the answer.

Answer: $x\in \left(-7;3 \right)$

That's all! Is it difficult? No, it's not difficult. Indeed, it was an easy task. Now let's complicate the mission a little and consider a more "fancy" inequality. When solving it, I will no longer give such detailed calculations - I will simply outline the key points. In general, we will arrange it the way we would have done it on an independent work or exam. :)

Task. Solve the inequality:

\[\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\ge 0$. All non-zero elements are collected on the left, there are no different denominators. Let's move on to equations.

Numerator:

\[\begin(align) & \left(7x+1 \right)\left(11x+2 \right)=0 \\ & 7x+1=0\Rightarrow ((x)_(1))=-\ frac(1)(7); \\ & 11x+2=0\Rightarrow ((x)_(2))=-\frac(2)(11). \\ \end(align)\]

Denominator:

\[\begin(align) & 13x-4=0; \\ & 13x=4; \\ & ((x)^(*))=\frac(4)(13). \\ \end(align)\]

I don’t know what kind of pervert made up this problem, but the roots didn’t turn out very well: it will be difficult to arrange them on a number line. And if everything is more or less clear with the root $((x)^(*))=(4)/(13)\;$ (this is the only positive number - it will be on the right), then $((x)_(1 ))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\;$ require further study: which one is larger?

You can find this out, for example:

\[((x)_(1))=-\frac(1)(7)=-\frac(2)(14) \gt -\frac(2)(11)=((x)_(2 ))\]

I hope there is no need to explain why the numeric fraction $-(2)/(14)\; \gt -(2)/(11)\;$? If necessary, I recommend remembering how to perform actions with fractions.

And we mark all three roots on the number line:

The points from the numerator are shaded, from the denominator they are cut out

We put up signs. For example, you can take $((x)_(0))=1$ and find out the sign at this point:

\[\begin(align) & f\left(x \right)=\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4); \\ & f\left(1 \right)=\frac(\left(7\cdot 1+1 \right)\left(11\cdot 1+2 \right))(13\cdot 1-4)=\ frac(8\cdot 13)(9) \gt 0. \\\end(align)\]

The last inequality before the equations was $f\left(x \right)\ge 0$, so we are interested in the plus sign.

We got two sets: one is an ordinary segment, and the other is an open ray on the number line.

Answer: $x\in \left[ -\frac(2)(11);-\frac(1)(7) \right]\bigcup \left(\frac(4)(13);+\infty \right )$

An important note about the numbers that we substitute to find out the sign on the rightmost interval. It is not necessary to substitute a number close to the rightmost root. You can take billions or even "plus-infinity" - in this case, the sign of the polynomial in the bracket, numerator or denominator is determined solely by the sign of the leading coefficient.

Let's take another look at the $f\left(x \right)$ function from the last inequality:

It contains three polynomials:

\[\begin(align) & ((P)_(1))\left(x \right)=7x+1; \\ & ((P)_(2))\left(x \right)=11x+2; \\ & Q\left(x\right)=13x-4. \end(align)\]

All of them are linear binomials, and all of them have positive coefficients (numbers 7, 11 and 13). Therefore, when substituting very large numbers, the polynomials themselves will also be positive. :)

This rule may seem overly complicated, but only at first, when we analyze very easy problems. In serious inequalities, the "plus-infinity" substitution will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.

We will face such challenges very soon. But first, let's look at an alternative way to solve fractional rational inequalities.

Alternative way

This technique was suggested to me by one of my students. I myself have never used it, but practice has shown that it is really more convenient for many students to solve inequalities in this way.

So, the original data is the same. We need to solve a fractional rational inequality:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\]

Let's think: why is the polynomial $Q\left(x \right)$ "worse" than the polynomial $P\left(x \right)$? Why do we have to consider separate groups of roots (with and without an asterisk), think about punched points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is different from zero.

Otherwise, there are no differences between the numerator and the denominator: we also equate it to zero, look for the roots, then mark them on the number line. So why not replace the fractional bar (in fact, the division sign) with the usual multiplication, and write all the requirements of the DHS as a separate inequality? For example, like this:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\Rightarrow \left\( \begin(align) & P\left(x \right)\cdot Q \left(x \right) \gt 0, \\ & Q\left(x \right)\ne 0. \\ \end(align) \right.\]

Please note: this approach will allow you to reduce the problem to the method of intervals, but it will not complicate the solution at all. After all, anyway, we will equate the polynomial $Q\left(x \right)$ to zero.

Let's see how it works on real tasks.

Task. Solve the inequality:

\[\frac(x+8)(x-11) \gt 0\]

Solution. So, let's move on to the interval method:

\[\frac(x+8)(x-11) \gt 0\Rightarrow \left\( \begin(align) & \left(x+8 \right)\left(x-11 \right) \gt 0 , \\ & x-11\ne 0. \\ \end(align) \right.\]

The first inequality is solved elementarily. Just set each parenthesis to zero:

\[\begin(align) & x+8=0\Rightarrow ((x)_(1))=-8; \\ & x-11=0\Rightarrow ((x)_(2))=11. \\ \end(align)\]

With the second inequality, everything is also simple:

We mark the points $((x)_(1))$ and $((x)_(2))$ on the real line. All of them are punctured because the inequality is strict:

The right point turned out to be punctured twice. This is fine.

Pay attention to the point $x=11$. It turns out that it is “twice gouged out”: on the one hand, we gouge it out because of the severity of inequality, on the other hand, because of the additional requirement of ODZ.

In any case, it will be just a punctured point. Therefore, we put signs for the inequality $\left(x+8 \right)\left(x-11 \right) \gt 0$ - the last one we saw before we started solving the equations:

We are interested in positive regions, since we are solving an inequality of the form $f\left(x \right) \gt 0$, and we will color them. It remains only to write down the answer.

Answer. $x\in \left(-\infty ;-8 \right)\bigcup \left(11;+\infty \right)$

Using this solution as an example, I would like to warn you against a common mistake among novice students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you a lot of problems.

Now let's try something more difficult.

Task. Solve the inequality:

\[\frac(\left(2x-13 \right)\left(12x-9 \right))(15x+33)\le 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\le 0$, so here you need to carefully monitor the filled points.

Let's move on to the interval method:

\[\left\( \begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)\le 0, \\ & 15x+33\ ne 0. \\ \end(align) \right.\]

Let's move on to the equation:

\[\begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0 \\ & 2x-13=0\Rightarrow ((x )_(1))=6.5; \\ & 12x-9=0\Rightarrow ((x)_(2))=0.75; \\ & 15x+33=0\Rightarrow ((x)_(3))=-2,2. \\ \end(align)\]

We take into account the additional requirement:

We mark all the obtained roots on the number line:

If a point is both punched out and filled in at the same time, it is considered punched out.

Again, two points "overlap" each other - this is normal, it will always be so. It is only important to understand that a point marked both as punched out and filled in is actually a punched out point. Those. "Gouging" is a stronger action than "painting over".

This is absolutely logical, because by puncturing we mark points that affect the sign of the function, but do not themselves participate in the answer. And if at some point the number ceases to suit us (for example, it does not fall into the ODZ), we delete it from consideration until the very end of the task.

In general, stop philosophizing. We arrange the signs and paint over those intervals that are marked with a minus sign:

Answer. $x\in \left(-\infty ;-2,2 \right)\bigcup \left[ 0,75;6,5 \right]$.

And again I wanted to draw your attention to this equation:

\[\left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0\]

Once again: never open parentheses in such equations! You're only making it harder for yourself. Remember: the product is zero when at least one of the factors is zero. Consequently, this equation simply “falls apart” into several smaller ones, which we solved in the previous problem.

Taking into account the multiplicity of roots

From the previous problems, it is easy to see that it is precisely the non-strict inequalities that are most difficult, because in them you have to keep track of the filled points.

But there is an even greater evil in the world - these are multiple roots in inequalities. Here it is already necessary to follow not some filled points there - here the inequality sign may not suddenly change when passing through these same points.

We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). So let's introduce a new definition:

Definition. The root of the equation $((\left(x-a \right))^(n))=0$ is equal to $x=a$ and is called the root of the $n$th multiplicity.

Actually, we are not particularly interested in the exact value of the multiplicity. The only important thing is whether this very number $n$ is even or odd. Because:

1. If $x=a$ is a root of even multiplicity, then the sign of the function does not change when passing through it;
2. And vice versa, if $x=a$ is a root of odd multiplicity, then the sign of the function will change.

A special case of a root of odd multiplicity are all the previous problems considered in this lesson: there the multiplicity is equal to one everywhere.

And further. Before we start solving problems, I would like to draw your attention to one subtlety that seems obvious to an experienced student, but drives many beginners into a stupor. Namely:

The multiplicity root $n$ arises only when the entire expression is raised to this power: $((\left(xa \right))^(n))$, and not $\left(((x)^( n))-a\right)$.

Once again: the bracket $((\left(xa \right))^(n))$ gives us the root $x=a$ of multiplicity $n$, but the bracket $\left(((x)^(n)) -a \right)$ or, as often happens, $(a-((x)^(n)))$ gives us a root (or two roots, if $n$ is even) of the first multiplicity, no matter what is equal to $n$.

Compare:

\[((\left(x-3 \right))^(5))=0\Rightarrow x=3\left(5k \right)\]

Everything is clear here: the whole bracket was raised to the fifth power, so at the output we got the root of the fifth degree. And now:

\[\left(((x)^(2))-4 \right)=0\Rightarrow ((x)^(2))=4\Rightarrow x=\pm 2\]

We got two roots, but both of them have the first multiplicity. Or here's another one:

\[\left(((x)^(10))-1024 \right)=0\Rightarrow ((x)^(10))=1024\Rightarrow x=\pm 2\]

And do not be confused by the tenth degree. The main thing is that 10 is an even number, so we have two roots at the output, and both of them again have the first multiplicity.

In general, be careful: multiplicity occurs only when the degree applies to the entire bracket, not just the variable.

Task. Solve the inequality:

\[\frac(((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right))(((\left(x+7 \right))^(5)))\ge 0\]

Solution. Let's try to solve it in an alternative way - through the transition from the particular to the product:

\[\left\( \begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ( (\left(x+7 \right))^(5))\ge 0, \\ & ((\left(x+7 \right))^(5))\ne 0. \\ \end(align )\right.\]

We deal with the first inequality using the interval method:

\[\begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ((\left( x+7 \right))^(5))=0; \\ & ((x)^(2))=0\Rightarrow x=0\left(2k \right); \\ & ((\left(6-x \right))^(3))=0\Rightarrow x=6\left(3k \right); \\ & x+4=0\Rightarrow x=-4; \\ & ((\left(x+7 \right))^(5))=0\Rightarrow x=-7\left(5k \right). \\ \end(align)\]

Additionally, we solve the second inequality. In fact, we have already solved it, but so that the reviewers do not find fault with the solution, it is better to solve it again:

\[((\left(x+7 \right))^(5))\ne 0\Rightarrow x\ne -7\]

Note that there are no multiplicities in the last inequality. Indeed: what difference does it make how many times to cross out the point $x=-7$ on the number line? At least once, at least five times - the result will be the same: a punctured point.

Let's note everything that we got on the number line:

As I said, the $x=-7$ point will eventually be punched out. The multiplicities are arranged based on the solution of the inequality by the interval method.

It remains to place the signs:

Since the point $x=0$ is a root of even multiplicity, the sign does not change when passing through it. The remaining points have an odd multiplicity, and everything is simple with them.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$

Pay attention to $x=0$ again. Because of the even multiplicity, an interesting effect arises: everything to the left of it is painted over, to the right - too, and the point itself is completely painted over.

As a consequence, it does not need to be isolated when recording a response. Those. you don't have to write something like $x\in \left[ -4;0 \right]\bigcup \left[ 0;6 \right]$ (although formally such an answer would also be correct). Instead, we immediately write $x\in \left[ -4;6 \right]$.

Such effects are possible only for roots of even multiplicity. And in the next task, we will encounter the reverse "manifestation" of this effect. Ready?

Task. Solve the inequality:

\[\frac(((\left(x-3 \right))^(4))\left(x-4 \right))(((\left(x-1 \right))^(2)) \left(7x-10-((x)^(2)) \right))\ge 0\]

Solution. This time we will follow the standard scheme. Set the numerator to zero:

\[\begin(align) & ((\left(x-3 \right))^(4))\left(x-4 \right)=0; \\ & ((\left(x-3 \right))^(4))=0\Rightarrow ((x)_(1))=3\left(4k \right); \\ & x-4=0\Rightarrow ((x)_(2))=4. \\ \end(align)\]

And the denominator:

\[\begin(align) & ((\left(x-1 \right))^(2))\left(7x-10-((x)^(2)) \right)=0; \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(1)^(*)=1\left(2k \right); \\ & 7x-10-((x)^(2))=0\Rightarrow x_(2)^(*)=5;\ x_(3)^(*)=2. \\ \end(align)\]

Since we are solving a non-strict inequality of the form $f\left(x \right)\ge 0$, the roots from the denominator (which have asterisks) will be cut out, and those from the numerator will be painted over.

We arrange the signs and stroke the areas marked with a "plus":

The point $x=3$ is isolated. This is part of the answer

Before writing down the final answer, take a close look at the picture:

1. The point $x=1$ has an even multiplicity, but is itself punctured. Therefore, it will have to be isolated in the answer: you need to write $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left(-\ infty ;2\right)$.
2. The point $x=3$ also has an even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step to the left and right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written as $x\in \left\( 3 \right\)$.

We combine all the obtained pieces into a common set and write down the answer.

Answer: $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;5 \right) $

Definition. Solving the inequality means find the set of all its solutions, or prove that this set is empty.

It would seem: what can be incomprehensible here? Yes, the fact of the matter is that sets can be specified in different ways. Let's rewrite the answer to the last problem:

We literally read what is written. The variable "x" belongs to a certain set, which is obtained by the union (symbol "U") of four separate sets:

• The interval $\left(-\infty ;1 \right)$, which literally means "all numbers less than one, but not one itself";
• The interval is $\left(1;2 \right)$, i.e. "all numbers between 1 and 2, but not the numbers 1 and 2 themselves";
• The set $\left\( 3 \right\)$, consisting of a single number - three;
• The interval $\left[ 4;5 \right)$ containing all numbers between 4 and 5, plus 4 itself, but not 5.

The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only denote only the boundaries of these sets, the set $\left\( 3 \right\)$ defines exactly one number by enumeration.

To understand that we are listing the specific numbers included in the set (and not setting boundaries or anything else), curly braces are used. For example, the notation $\left\( 1;2 \right\)$ means exactly "a set consisting of two numbers: 1 and 2", but not a segment from 1 to 2. In no case do not confuse these concepts.

Multiplicity addition rule

Well, at the end of today's lesson, a little tin from Pavel Berdov. :)

Attentive students have probably already asked themselves the question: what will happen if the same roots are found in the numerator and denominator? So the following rule works:

Multiplicities of identical roots are added. Is always. Even if this root occurs in both the numerator and the denominator.

Sometimes it's better to decide than to talk. Therefore, we solve the following problem:

Task. Solve the inequality:

\[\frac(((x)^(2))+6x+8)(\left(((x)^(2))-16 \right)\left(((x)^(2))+ 9x+14 \right))\ge 0\]

\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -4. \\ \end(align)\]

So far, nothing special. Set the denominator to zero:

\[\begin(align) & \left(((x)^(2))-16 \right)\left(((x)^(2))+9x+14 \right)=0 \\ & ( (x)^(2))-16=0\Rightarrow x_(1)^(*)=4;\ x_(2)^(*)=-4; \\ & ((x)^(2))+9x+14=0\Rightarrow x_(3)^(*)=-7;\ x_(4)^(*)=-2. \\ \end(align)\]

Two identical roots are found: $((x)_(1))=-2$ and $x_(4)^(*)=-2$. Both have the first multiplicity. Therefore, we replace them with one root $x_(4)^(*)=-2$, but with a multiplicity of 1+1=2.

In addition, there are also identical roots: $((x)_(2))=-4$ and $x_(2)^(*)=-4$. They are also of the first multiplicity, so only $x_(2)^(*)=-4$ of multiplicity 1+1=2 remains.

Please note: in both cases, we left exactly the “cut out” root, and threw out the “painted over” one from consideration. Because even at the beginning of the lesson, we agreed: if a point is both punched out and painted over at the same time, then we still consider it punched out.

As a result, we have four roots, and all of them turned out to be gouged out:

\[\begin(align) & x_(1)^(*)=4; \\ & x_(2)^(*)=-4\left(2k \right); \\ & x_(3)^(*)=-7; \\ & x_(4)^(*)=-2\left(2k \right). \\ \end(align)\]

We mark them on the number line, taking into account the multiplicity:

We place the signs and paint over the areas of interest to us:

Everything. No isolated points and other perversions. You can write down the answer.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left(4;+\infty \right)$.

multiplication rule

Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to a certain power. This changes the multiplicities of all the original roots.

This is rare, so most students do not have experience in solving such problems. And the rule here is:

When an equation is raised to a power $n$, the multiplicity of all its roots also increases by a factor of $n$.

In other words, raising to a power results in multiplying multiplicities by the same power. Let's take this rule as an example:

Task. Solve the inequality:

\[\frac(x((\left(((x)^(2))-6x+9 \right))^(2))((\left(x-4 \right))^(5)) )(((\left(2-x \right))^(3))((\left(x-1 \right))^(2)))\le 0\]

Solution. Set the numerator to zero:

The product is equal to zero when at least one of the factors is equal to zero. Everything is clear with the first multiplier: $x=0$. And here's where the problems start:

\[\begin(align) & ((\left(((x)^(2))-6x+9 \right))^(2))=0; \\ & ((x)^(2))-6x+9=0\left(2k \right); \\ & D=((6)^(3))-4\cdot 9=0 \\ & ((x)_(2))=3\left(2k \right)\left(2k \right) \ \ & ((x)_(2))=3\left(4k \right) \\ \end(align)\]

As you can see, the equation $((x)^(2))-6x+9=0$ has a unique root of the second multiplicity: $x=3$. The whole equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which we finally wrote down.

\[((\left(x-4 \right))^(5))=0\Rightarrow x=4\left(5k \right)\]

No problem with the denominator either:

\[\begin(align) & ((\left(2-x \right))^(3))((\left(x-1 \right))^(2))=0; \\ & ((\left(2-x \right))^(3))=0\Rightarrow x_(1)^(*)=2\left(3k \right); \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(2)^(*)=1\left(2k \right). \\ \end(align)\]

In total, we got five points: two punched out and three filled in. There are no coinciding roots in the numerator and denominator, so we just mark them on the number line:

We arrange the signs taking into account the multiplicities and paint over the intervals of interest to us:

Again one isolated point and one punctured

Because of the roots of even multiplicity, we again received a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, not $x\in \left[ 0;2 \right)$, and also an isolated point $ x\in \left\( 3 \right\)$.

Answer. $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;+\infty \right)$

As you can see, everything is not so difficult. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the very ones that we discussed at the very beginning.

Preconversions

The inequalities we will discuss in this section are not complex. However, unlike the previous tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.

We discussed this issue in detail at the very beginning of today's lesson. If you are not sure that you understand what it is about, I strongly recommend that you go back and repeat. Because there is no point in cramming the methods for solving inequalities if you "swim" in the conversion of fractions.

In homework, by the way, there will also be many similar tasks. They are placed in a separate subsection. And there you will find very non-trivial examples. But this will be in the homework, but now let's analyze a couple of such inequalities.

Task. Solve the inequality:

\[\frac(x)(x-1)\le \frac(x-2)(x)\]

Solution. Moving everything to the left:

\[\frac(x)(x-1)-\frac(x-2)(x)\le 0\]

We reduce to a common denominator, open the brackets, give like terms in the numerator:

\[\begin(align) & \frac(x\cdot x)(\left(x-1 \right)\cdot x)-\frac(\left(x-2 \right)\left(x-1 \ right))(x\cdot \left(x-1 \right))\le 0; \\ & \frac(((x)^(2))-\left(((x)^(2))-2x-x+2 \right))(x\left(x-1 \right)) \le0; \\ & \frac(((x)^(2))-((x)^(2))+3x-2)(x\left(x-1 \right))\le 0; \\ & \frac(3x-2)(x\left(x-1 \right))\le 0. \\\end(align)\]

Now we have a classical fractional rational inequality, the solution of which is no longer difficult. I propose to solve it by an alternative method - through the method of intervals:

\[\begin(align) & \left(3x-2 \right)\cdot x\cdot \left(x-1 \right)=0; \\ & ((x)_(1))=\frac(2)(3);\ ((x)_(2))=0;\ ((x)_(3))=1. \\ \end(align)\]

Don't forget the constraint that comes from the denominator:

We mark all the numbers and restrictions on the number line:

All roots have first multiplicity. No problem. We just place the signs and paint over the areas we need:

It's all. You can write down the answer.

Answer. $x\in \left(-\infty ;0 \right)\bigcup \left[ (2)/(3)\;;1 \right)$.

Of course, this was a very simple example. So now let's take a closer look at the problem. And by the way, the level of this task is quite consistent with independent and control work on this topic in the 8th grade.

Task. Solve the inequality:

\[\frac(1)(((x)^(2))+8x-9)\ge \frac(1)(3((x)^(2))-5x+2)\]

Solution. Moving everything to the left:

\[\frac(1)(((x)^(2))+8x-9)-\frac(1)(3((x)^(2))-5x+2)\ge 0\]

Before bringing both fractions to a common denominator, we decompose these denominators into factors. Suddenly the same brackets will come out? With the first denominator it's easy:

\[((x)^(2))+8x-9=\left(x-1 \right)\left(x+9 \right)\]

The second one is a little more difficult. Feel free to add a constant multiplier to the bracket where the fraction was found. Remember: the original polynomial had integer coefficients, so it is highly likely that the factorization will also have integer coefficients (in fact, it always will, except when the discriminant is irrational).

\[\begin(align) & 3((x)^(2))-5x+2=3\left(x-1 \right)\left(x-\frac(2)(3) \right)= \\ & =\left(x-1 \right)\left(3x-2 \right) \end(align)\]

As you can see, there is a common bracket: $\left(x-1 \right)$. We return to the inequality and bring both fractions to a common denominator:

\[\begin(align) & \frac(1)(\left(x-1 \right)\left(x+9 \right))-\frac(1)(\left(x-1 \right)\ left(3x-2\right))\ge 0; \\ & \frac(1\cdot \left(3x-2 \right)-1\cdot \left(x+9 \right))(\left(x-1 \right)\left(x+9 \right )\left(3x-2 \right))\ge 0; \\ & \frac(3x-2-x-9)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ & \frac(2x-11)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ \end(align)\]

Set the denominator to zero:

\[\begin(align) & \left(x-1 \right)\left(x+9 \right)\left(3x-2 \right)=0; \\ & x_(1)^(*)=1;\ x_(2)^(*)=-9;\ x_(3)^(*)=\frac(2)(3) \\ \end( align)\]

No multiplicities and no coinciding roots. We mark four numbers on a straight line:

We place the signs:

We write down the answer.

Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5,5;+\infty \ right)$.

Spacing Method- this is a universal way to solve almost any inequalities that occur in a school algebra course. It is based on the following properties of functions:

1. The continuous function g(x) can change sign only at the point where it is equal to 0. Graphically, this means that the graph of a continuous function can move from one half-plane to another only if it crosses the x-axis (we remember that the ordinate of any point lying on the OX axis (abscissa axis) is equal to zero, that is, the value of the function at this point is 0):

We see that the function y=g(x) shown on the graph crosses the OX axis at the points x= -8, x=-2, x=4, x=8. These points are called zeros of the function. And at the same points the function g(x) changes sign.

2. The function can also change the sign at zeros of the denominator - the simplest example of a well-known function:

We see that the function changes sign at the root of the denominator, at the point , but does not vanish at any point. Thus, if the function contains a fraction, it can change the sign in the roots of the denominator.

2. However, the function does not always change sign at the root of the numerator or at the root of the denominator. For example, the function y=x 2 does not change sign at the point x=0:

Because the equation x 2 \u003d 0 has two equal roots x \u003d 0, at the point x \u003d 0, the function, as it were, turns to 0 twice. Such a root is called the root of the second multiplicity.

Function changes sign at zero of the numerator, but does not change sign at zero of the denominator: , since the root is the root of the second multiplicity, that is, of even multiplicity:

Important! At roots of even multiplicity, the function does not change sign.

Note! Any non-linear the inequality of the school course of algebra, as a rule, is solved using the method of intervals.

I offer you a detailed one, following which you can avoid mistakes when solving nonlinear inequalities.

1. First you need to bring the inequality to the form

P(x)V0,

where V is the inequality sign:<,>,≤ or ≥. For this you need:

a) move all the terms to the left side of the inequality,

b) find the roots of the resulting expression,

c) factorize the left side of the inequality

d) write the same factors as a degree.

Attention! The last action must be done in order not to make a mistake with the multiplicity of the roots - if the result is a multiplier in an even degree, then the corresponding root has an even multiplicity.

2. Put the found roots on the number line.

3. If the inequality is strict, then the circles denoting the roots on the numerical axis are left "empty", if the inequality is not strict, then the circles are painted over.

4. We select the roots of even multiplicity - in them P(x) the sign does not change.

5. Determine the sign P(x) on the right side of the gap. To do this, take an arbitrary value x 0, which is greater than the largest root and substitute in P(x).

If P(x 0)>0 (or ≥0), then in the rightmost interval we put the "+" sign.

If P(x0)<0 (или ≤0), то в самом правом промежутке ставим знак "-".

When passing through a point denoting a root of even multiplicity, the sign DOES NOT change.

7. Once again we look at the sign of the original inequality, and select the intervals of the sign we need.

8. Attention! If our inequality is NOT STRICT, then we check the condition of equality to zero separately.

9. Write down the answer.

If the original the inequality contains an unknown in the denominator, then we also transfer all the terms to the left, and reduce the left side of the inequality to the form

(where V is the inequality sign:< или >)

A strict inequality of this kind is equivalent to the inequality

NOT strict an inequality of the form

is tantamount to system:

In practice, if the function has the form , then we proceed as follows:

1. Find the roots of the numerator and denominator.
2. We put them on the axis. All circles are left empty. Then, if the inequality is not strict, then we paint over the roots of the numerator, and always leave the roots of the denominator empty.
3. Next, we follow the general algorithm:
4. We select the roots of even multiplicity (if the numerator and denominator contain the same roots, then we count how many times the same roots occur). There is no change of sign in roots of even multiplicity.
5. We find out the sign on the rightmost interval.
6. We put up signs.
7. In the case of a nonstrict inequality, the condition of equality, the condition of equality to zero, is checked separately.
8. We select the necessary intervals and separately standing roots.
9. We write down the answer.

To better understand algorithm for solving inequalities by the interval method, watch the VIDEO LESSON in which the example is analyzed in detail solution of inequality by the method of intervals.

First level

interval method. Comprehensive 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 will reveal to you a couple of secrets on how to save time on solving these inequalities. Well, are you intrigued? Then let's go!

The essence of the method is to factorize the inequality (repeat the topic) and determine the ODZ and the sign of the factors, now I’ll explain everything. Let's take the simplest example: .

It is not necessary to write the area of ​​​​admissible values ​​() here, since there is no division by a variable, and radicals (roots) are not observed here. Everything here is already multiplied for us. But do not relax, this is all to remind the basics and understand the essence!

Suppose you do not know the method of intervals, how would you go about solving this inequality? Be logical and build on what you already know. First, the left side will be greater than zero if both parenthesized expressions are either greater than zero or less than zero, since "Plus" on "plus" makes "plus" and "minus" on "minus" makes "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. But what do we need to find out those values ​​for which the expressions in brackets will be negative or positive?

We need to solve the equation, it is exactly the same as the inequality, only instead of the sign there will be a sign, the roots of this equation will allow us to determine those boundary values, deviating from which the factors and 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 enclosed between some two numbers - the ends of the interval. It’s not so easy to imagine these gaps in your head, so it’s customary to draw intervals, now I’ll teach you.

We draw an axis, on it the entire number series is located from and to. Points are plotted on the axis, the very so-called zeros of the function, values ​​at which the expression equals zero. These points are "pricked out" which means that they are not among those values ​​for which the inequality is true. In this case, they are punctured. the 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 without circles here, but so, for understanding and orientation along the axis. Okay, the axis was drawn, the dots (or rather circles) were set, then what, 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 the sign will be as a result of multiplication.

In short, we just take an example, substitute it here, it will turn out, which means that on the entire interval (on the entire interval) from to, from which we took, the inequality will be true. In other words, if x is from to, then the inequality is true.

We do the same with an 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 will turn out to be “plus”. Such a bunch of text came out, but there is little visibility, right?

Look again at inequality.

Now, on the same axis, we also apply the signs that will be the result. The broken line, in my example, denotes the positive and negative sections of the axis.

Look at the inequality - at the picture, again at the inequality - and again at the picture is anything clear? Now try to say on what intervals of x, the inequality will be true. That's right, from to the inequality will also be true from to, and on the interval from to the inequality of zero and this interval is of little interest to us, because we have a sign in the inequality.

Well, since you figured it out, then it’s up to you to write down the answer! In response, we write those intervals at which the left side is greater than zero, which is read 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 ​​that the interval is bounded by are not solutions to the inequality, that is, they are not included in the answer, but only say that before, for example, but there is no solution.

Now an example in which you will have to draw not only the interval:

What do you think should be done before putting points on the axis? Yep, factor it out:

We draw intervals and place signs, notice the points we have punctured, because the sign is strictly less than zero:

It's time to reveal to you one secret that I promised at the beginning of this topic! But what if I tell you that you can not substitute the values ​​from each interval to determine the sign, but you can determine the sign in one of the intervals, and in the rest just alternate the signs!

Thus, we saved a little time on putting down signs - I think this time won on the 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 there’s a fraction, and even with an unknown in the denominator!

That's right, ODZ is necessary!

So, let's go further, here all 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 should be x in each factor. And here, so it is always positive, because. any squared number > zero and a positive term.

How do you think it will affect the value of inequality? That's right - it doesn't matter! We can safely divide the inequality into both parts and thereby remove this factor so that it does not hurt our eyes.

it's time to draw intervals, for this you need to determine those boundary values, deviating from which the multipliers and will be greater and less than zero. But pay attention that here the sign means the point at which the left side of the inequality takes on a zero value, we will not puncture it, because it is included in the number of solutions, we have one such point, this is the point where x is equal to one. Can 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:

According to this scheme, you can already easily write an answer, I can only say that now you have a new type of bracket at your disposal - square! Here is a bracket [ says that the value is in the solution interval, i.e. is part of the answer, this parenthesis corresponds to a filled (not punched out) point on the axis.

So, did you get the same answer?

We factorize and transfer everything in one direction, because 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 get in the numerator as well as in the denominator, I multiply both parts of the inequality by. Remember that when you multiply both sides of the inequality by, the sign of the inequality is reversed!!!

We write ODZ:

Otherwise, the denominator will turn to zero, and, as you remember, you cannot divide by zero!

Agree, in the resulting inequality it is tempting to reduce in the numerator and denominator! You can’t do this, you can lose some of the decisions or ODZ!

Now try to put points on the axis yourself. I will only note that when drawing points, you need to pay attention to the fact that a point with a value, which, based on the sign, it would seem, should be drawn on the axis as filled in, will not be filled in, it will be punched out! Why ask you? And you remember ODZ, you're not going to divide by zero like that?

Remember, ODZ is above all! If all inequality and equal signs say one thing, and the ODZ says another, trust the ODZ, great and mighty! Well, you built the intervals, I'm sure you took my tip about alternation and you got it like this (see the picture below) Now cross it out and don't repeat this mistake again! What mistake? - you ask.

The fact is that in this inequality the factor was repeated twice (remember how you still 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 odd, then the sign changes!

The following axis with intervals and signs will be correct:

And, pay attention that the sign we are not interested in is the one that was at the beginning (when we just saw the inequality, the sign was), after the transformations, the sign changed to, which means that we are interested in the intervals with the sign.

Answer:

I will also say that there are situations when there are roots of inequality that are not included in any gap, in response they are written in curly brackets, like this, for example:. You can read more about such situations in the article Intermediate Level.

Let's summarize how to solve inequalities using the interval method:

1. We transfer everything to the left side, on the right we leave only zero;
2. We find ODZ;
3. We put on the axis all the roots of the inequality;
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, it depends on the even or odd number of times of their repetition whether the sign changes when passing through them or not;
5. In response, we write the intervals, observing the punched out and not punched out points (see ODZ), putting 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 real axis:

We say that "the function changes sign when passing through a point".

It can be seen that the signs of the function correspond to the position of the graph of the function: if the graph is above the axis, the sign is “ ”, if it is below - “ ”.

If we generalize the resulting rule to an arbitrary linear function, we get the following algorithm:

• We find the zero of the function;
• We mark it on the numerical axis;
• We determine the sign of the function on opposite sides of zero.

quadratic function

I hope you remember how quadratic inequalities are solved? If not, read the thread. Let me remind you 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 the 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 change alternately when passing through each root.

Is it possible to somehow determine the signs without drawing a parabola each time?

Recall that the square trinomial can be factorized:

For instance: .

Note the roots on the axis:

We remember that the sign of a function can only change when passing through the root. We 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 only at one arbitrarily chosen point: at the other points of the interval, the sign will be the same.

In our example: for both expressions in brackets are positive (we substitute, for example:). We put the sign "" on the axis:

Well, if (substitute, for example) both brackets are negative, then 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 us also consider cases when the function has no zeros, or it is only one.

If there are none, then there are no roots. This means that there will be no “passage through the root”. This means that the function on the entire number axis takes only one sign. It is easy to determine 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 is the rule for such situations?

If we factor out such a function, we 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 select the root, when passing through which the sign does not change, circling it with a square:

Such a root will be called a multiple.

The method of intervals 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 choose the intervals depending on the inequality sign. For instance:

We measure the roots on the axis and arrange 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 parts of which are rational expressions (see).

Example:

All factors except one - - here are "linear", that is, they contain a variable only in the first degree. 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 it yourself), and therefore does not affect the sign of the entire inequality. This means that you can divide the left and right sides of the inequality into 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 vanishes, mark these points on the axis and arrange the signs. I draw your attention to a very important fact:

Answer: . Example: .

To apply the interval method, it is necessary that in one of the parts of the inequality was. Therefore, we move the right side to the left:

The numerator and denominator have the same factor, but we are not in a hurry to reduce it! After all, then we can forget to poke 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 another very illustrative example:

Again, we do not reduce the same factors of the numerator and denominator, because if we reduce, we will have to specifically remember that we need to poke a point.

• : repeated times;
• : times;
• : times (in the numerator and one in the denominator).

In the case of an even number, we proceed in the same way as before: we circle the point with a square and do not change the sign when passing through the root. But in the case of an odd number, this rule is not fulfilled: the sign will still change when passing through the root. Therefore, we do nothing additionally with such a root, as if it is not a multiple of us. The above rules apply to all even and odd powers.

What do we write in the answer?

If the alternation of signs is violated, you need to be very careful, because with a non-strict inequality, the answer should include all filled points. But some of them often stand alone, that is, they do not enter the shaded area. In this case, we add them to the response as isolated dots (in curly braces):

Examples (decide for yourself):

Answers:

1. If among the factors it is simple - this is the root, because it can be represented as.
   .

INTERVAL METHOD. BRIEFLY ABOUT THE MAIN

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 different intervals.

Algorithm for solving rational inequalities by the interval method.

• We transfer everything to the left side, on the right we leave only zero;
• We find ODZ;
• We put on the axis all the roots of the inequality;
• 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, it depends on the even or odd number of times of their repetition whether the sign changes when passing through them or not;
• In response, we write intervals, observing the punched out and not punched out points (see ODZ), putting the necessary types of brackets between them.

Well, the topic is over. If you are reading these lines, then you are very cool.

Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!

Now the most important thing.

You've figured out the theory on this topic. And, I repeat, it's ... it's 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 the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.

I will not convince you of anything, I will 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 much more opportunities open up before them and life becomes brighter? Do not know...

But think for yourself...

What does it take to be sure to be better than others on the exam and be ultimately ... happier?

FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.

On the exam, you will not be asked theory.

You will need solve problems on time.

And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.

It's like in sports - you need to repeat many times to win for sure.

Find a collection anywhere you want necessarily with solutions, detailed analysis and decide, decide, decide!

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Find problems and solve!

First level

interval method. Comprehensive 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 will reveal to you a couple of secrets on how to save time on solving these inequalities. Well, are you intrigued? Then let's go!

The essence of the method is to factorize the inequality (repeat the topic) and determine the ODZ and the sign of the factors, now I’ll explain everything. Let's take the simplest example: .

It is not necessary to write the area of ​​​​admissible values ​​() here, since there is no division by a variable, and radicals (roots) are not observed here. Everything here is already multiplied for us. But do not relax, this is all to remind the basics and understand the essence!

Suppose you do not know the method of intervals, how would you go about solving this inequality? Be logical and build on what you already know. First, the left side will be greater than zero if both parenthesized expressions are either greater than zero or less than zero, since "Plus" on "plus" makes "plus" and "minus" on "minus" makes "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. But what do we need to find out those values ​​for which the expressions in brackets will be negative or positive?

We need to solve the equation, it is exactly the same as the inequality, only instead of the sign there will be a sign, the roots of this equation will allow us to determine those boundary values, deviating from which the factors and 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 enclosed between some two numbers - the ends of the interval. It’s not so easy to imagine these gaps in your head, so it’s customary to draw intervals, now I’ll teach you.

We draw an axis, on it the entire number series is located from and to. Points are plotted on the axis, the very so-called zeros of the function, values ​​at which the expression equals zero. These points are "pricked out" which means that they are not among those values ​​for which the inequality is true. In this case, they are punctured. the 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 without circles here, but so, for understanding and orientation along the axis. Okay, the axis was drawn, the dots (or rather circles) were set, then what, 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 the sign will be as a result of multiplication.

In short, we just take an example, substitute it here, it will turn out, which means that on the entire interval (on the entire interval) from to, from which we took, the inequality will be true. In other words, if x is from to, then the inequality is true.

We do the same with an 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 will turn out to be “plus”. Such a bunch of text came out, but there is little visibility, right?

Look again at inequality.

Now, on the same axis, we also apply the signs that will be the result. The broken line, in my example, denotes the positive and negative sections of the axis.

Look at the inequality - at the picture, again at the inequality - and again at the picture is anything clear? Now try to say on what intervals of x, the inequality will be true. That's right, from to the inequality will also be true from to, and on the interval from to the inequality of zero and this interval is of little interest to us, because we have a sign in the inequality.

Well, since you figured it out, then it’s up to you to write down the answer! In response, we write those intervals at which the left side is greater than zero, which is read 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 ​​that the interval is bounded by are not solutions to the inequality, that is, they are not included in the answer, but only say that before, for example, but there is no solution.

Now an example in which you will have to draw not only the interval:

What do you think should be done before putting points on the axis? Yep, factor it out:

We draw intervals and place signs, notice the points we have punctured, because the sign is strictly less than zero:

It's time to reveal to you one secret that I promised at the beginning of this topic! But what if I tell you that you can not substitute the values ​​from each interval to determine the sign, but you can determine the sign in one of the intervals, and in the rest just alternate the signs!

Thus, we saved a little time on putting down signs - I think this time won on the 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 there’s a fraction, and even with an unknown in the denominator!

That's right, ODZ is necessary!

So, let's go further, here all 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 should be x in each factor. And here, so it is always positive, because. any squared number > zero and a positive term.

How do you think it will affect the value of inequality? That's right - it doesn't matter! We can safely divide the inequality into both parts and thereby remove this factor so that it does not hurt our eyes.

it's time to draw intervals, for this you need to determine those boundary values, deviating from which the multipliers and will be greater and less than zero. But pay attention that here the sign means the point at which the left side of the inequality takes on a zero value, we will not puncture it, because it is included in the number of solutions, we have one such point, this is the point where x is equal to one. Can 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:

According to this scheme, you can already easily write an answer, I can only say that now you have a new type of bracket at your disposal - square! Here is a bracket [ says that the value is in the solution interval, i.e. is part of the answer, this parenthesis corresponds to a filled (not punched out) point on the axis.

So, did you get the same answer?

We factorize and transfer everything in one direction, because 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 get in the numerator as well as in the denominator, I multiply both parts of the inequality by. Remember that when you multiply both sides of the inequality by, the sign of the inequality is reversed!!!

We write ODZ:

Otherwise, the denominator will turn to zero, and, as you remember, you cannot divide by zero!

Agree, in the resulting inequality it is tempting to reduce in the numerator and denominator! You can’t do this, you can lose some of the decisions or ODZ!

Now try to put points on the axis yourself. I will only note that when drawing points, you need to pay attention to the fact that a point with a value, which, based on the sign, it would seem, should be drawn on the axis as filled in, will not be filled in, it will be punched out! Why ask you? And you remember ODZ, you're not going to divide by zero like that?

Remember, ODZ is above all! If all inequality and equal signs say one thing, and the ODZ says another, trust the ODZ, great and mighty! Well, you built the intervals, I'm sure you took my tip about alternation and you got it like this (see the picture below) Now cross it out and don't repeat this mistake again! What mistake? - you ask.

The fact is that in this inequality the factor was repeated twice (remember how you still 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 odd, then the sign changes!

The following axis with intervals and signs will be correct:

And, pay attention that the sign we are not interested in is the one that was at the beginning (when we just saw the inequality, the sign was), after the transformations, the sign changed to, which means that we are interested in the intervals with the sign.

Answer:

I will also say that there are situations when there are roots of inequality that are not included in any gap, in response they are written in curly brackets, like this, for example:. You can read more about such situations in the article Intermediate Level.

Let's summarize how to solve inequalities using the interval method:

1. We transfer everything to the left side, on the right we leave only zero;
2. We find ODZ;
3. We put on the axis all the roots of the inequality;
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, it depends on the even or odd number of times of their repetition whether the sign changes when passing through them or not;
5. In response, we write the intervals, observing the punched out and not punched out points (see ODZ), putting 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 real axis:

We say that "the function changes sign when passing through a point".

It can be seen that the signs of the function correspond to the position of the graph of the function: if the graph is above the axis, the sign is “ ”, if it is below - “ ”.

If we generalize the resulting rule to an arbitrary linear function, we get the following algorithm:

• We find the zero of the function;
• We mark it on the numerical axis;
• We determine the sign of the function on opposite sides of zero.

quadratic function

I hope you remember how quadratic inequalities are solved? If not, read the thread. Let me remind you 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 the 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 change alternately when passing through each root.

Is it possible to somehow determine the signs without drawing a parabola each time?

Recall that the square trinomial can be factorized:

For instance: .

Note the roots on the axis:

We remember that the sign of a function can only change when passing through the root. We 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 only at one arbitrarily chosen point: at the other points of the interval, the sign will be the same.

In our example: for both expressions in brackets are positive (we substitute, for example:). We put the sign "" on the axis:

Well, if (substitute, for example) both brackets are negative, then 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 us also consider cases when the function has no zeros, or it is only one.

If there are none, then there are no roots. This means that there will be no “passage through the root”. This means that the function on the entire number axis takes only one sign. It is easy to determine 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 is the rule for such situations?

If we factor out such a function, we 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 select the root, when passing through which the sign does not change, circling it with a square:

Such a root will be called a multiple.

The method of intervals 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 choose the intervals depending on the inequality sign. For instance:

We measure the roots on the axis and arrange 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 parts of which are rational expressions (see).

Example:

All factors except one - - here are "linear", that is, they contain a variable only in the first degree. 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 it yourself), and therefore does not affect the sign of the entire inequality. This means that you can divide the left and right sides of the inequality into 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 vanishes, mark these points on the axis and arrange the signs. I draw your attention to a very important fact:

Answer: . Example: .

To apply the interval method, it is necessary that in one of the parts of the inequality was. Therefore, we move the right side to the left:

The numerator and denominator have the same factor, but we are not in a hurry to reduce it! After all, then we can forget to poke 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 another very illustrative example:

Again, we do not reduce the same factors of the numerator and denominator, because if we reduce, we will have to specifically remember that we need to poke a point.

• : repeated times;
• : times;
• : times (in the numerator and one in the denominator).

In the case of an even number, we proceed in the same way as before: we circle the point with a square and do not change the sign when passing through the root. But in the case of an odd number, this rule is not fulfilled: the sign will still change when passing through the root. Therefore, we do nothing additionally with such a root, as if it is not a multiple of us. The above rules apply to all even and odd powers.

What do we write in the answer?

If the alternation of signs is violated, you need to be very careful, because with a non-strict inequality, the answer should include all filled points. But some of them often stand alone, that is, they do not enter the shaded area. In this case, we add them to the response as isolated dots (in curly braces):

Examples (decide for yourself):

Answers:

1. If among the factors it is simple - this is the root, because it can be represented as.
   .

INTERVAL METHOD. BRIEFLY ABOUT THE MAIN

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 different intervals.

Algorithm for solving rational inequalities by the interval method.

• We transfer everything to the left side, on the right we leave only zero;
• We find ODZ;
• We put on the axis all the roots of the inequality;
• 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, it depends on the even or odd number of times of their repetition whether the sign changes when passing through them or not;
• In response, we write intervals, observing the punched out and not punched out points (see ODZ), putting the necessary types of brackets between them.

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