Solution of inequalities by the method of intervals. Solving quadratic inequalities by the interval method

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

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

We pay special attention to the nuances of the topic, which are usually not covered in the school curriculum. For example, consider the rules for placing signs on intervals and the method of intervals in general view without linking it to rational inequalities.

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Algorithm

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

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

Here are some examples of such inequalities:

(x + 3) (x 2 − x + 1) (x + 2) 3 ≥ 0,

(x - 2) (x + 5) x + 3 > 0 ,

(x − 5) (x + 5) ≤ 0 ,

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

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

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

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

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

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

Scientific basis of the gap method

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

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

It is also possible to justify the constancy of the sign on the intervals on the basis of the properties of numerical inequalities. For example, take the inequality x - 5 x + 1 > 0 . If we find the zeros of the numerator and denominator and put them on the number line, we get a series of gaps: (− ∞ , − 1) , (− 1 , 5) and (5 , + ∞) .

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

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

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

Finding zeros of the numerator and denominator

The algorithm for finding zeros is simple: we equate the expressions from the numerator and denominator to zero and solve the resulting equations. If you have any difficulties, you can refer to the topic "Solving Equations by Factoring". In this section, we confine ourselves to an example.

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

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

The second equation is equivalent to the set of three equations x7 = 0, (x 2 + 2 x + 7) 2 = 0, (x + 5) 3 = 0 . We carry out a series of transformations and get x \u003d 0, x 2 + 2 x + 7 \u003d 0, x + 5 \u003d 0. The root of the first equation is 0, the second equation has no roots, since it has a negative discriminant, the root of the third equation is 5. These are the zeros of the denominator.

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

In general, when there is a fraction on the left side of the inequality, which is not necessarily rational, the numerator and denominator are also equated to zero to obtain equations. Solving equations allows you to find the zeros of the numerator and denominator.

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

Let's look at this statement with an example.

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

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

From the first interval (− ∞ , − 3) take - 4 . At x = -4 we have (- 4) 2 - (- 4) + 4 (- 4) + 3 = - 24 . We got negative meaning, so the entire interval will be with the sign "-".

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

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

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

Now let's turn to examples.

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

The zeros of the denominator are marked with empty dots.

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

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

Moving from right to left, we will mark the remaining gaps. We pass through the point with coordinate 4 . It is both the zero of the numerator and the denominator. In sum, these zeros give the expressions (x − 4) 2 and x − 4. We add their powers 2 + 1 = 3 and get odd number. This means that the sign in the transition in this case changes to the opposite. On the interval (3, 4) there will be a minus sign.

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

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

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

The use of the interval method is especially effective in cases where the calculation of the value of an expression is associated with a large amount of work. An example would be the need to evaluate the value of an expression

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

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

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

Example 1

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

Solution

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

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

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

Answer: (- ∞ , 1) ∪ (1 , 7) .

The solution of rational inequalities in most cases requires their preliminary transformation to the right kind. Only then does it become possible to use the interval method. Algorithms for carrying out such transformations are considered in the material "Solution of rational inequalities".

Consider an example of converting square trinomials into inequalities.

Example 2

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

Solution

Let's see if the discriminants of square trinomials in the inequality record are really negative. This will allow us to determine whether the form of this inequality allows us to apply the interval method to the solution.

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

Answer: (- 4 , - 3) ∪ (2 , + ∞) .

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

All actions are carried out according to a certain algorithm. In this case, the algorithm for solving inequalities by the generalized interval method will differ somewhat from what we have analyzed earlier:

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

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

The zeros of the function in the case of non-strict inequalities are marked with ordinary (shaded) dots, and for strict inequalities, with empty dots. If the zeros coincide with the boundary points or individual points of the domain of definition, then they can be recolored in black, making them empty or filled, depending on the type of inequality.

The response record is number set which includes:

hatched gaps;
individual points of the domain with a plus sign if we are dealing with an inequality whose sign is > or ≥ or with a minus sign if there are signs in the inequality< или ≤ .

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

Consider an example of applying the generalized interval method.

Example 3

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

Solution

We introduce a function f such that f (x) = x 2 + 2 x - 24 - 3 4 x - 3 x - 7 . Find the domain of the function f:

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

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

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

We get the root x = 12 .

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

We mark the zero of the function with an empty black dot, since we are working with strict inequality.

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

f (16) = 16 2 + 2 16 - 24 - 3 4 16 - 3 16 - 7 = 264 - 15 9 > 0 f (8) = 8 2 + 2 8 - 24 - 3 4 8 - 3 8 - 7 = 56 - 9< 0 f (6) = 6 2 + 2 · 6 - 24 - 3 4 · 6 - 3 6 - 7 = 24 - 15 2 - 1 = = 15 - 2 · 24 2 = 225 - 96 2 >0 f (- 8) \u003d - 8 2 + 2 (- 8) - 24 - 3 4 (- 8) - 3 - 8 - 7 \u003d 24 + 3 - 15< 0

We place the signs we just defined, and we apply hatching over the gaps with a minus sign:

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

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

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

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

these dots are part of a hatched gap,
these points are separate points of the domain of the function, the values of the function in which satisfy the inequality being solved.

Answer: (− ∞ , − 6 ] ∪ (7 , 12) .

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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 to factorize square trinomial, that is, an expression of the form .

Where and are the roots 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. Consider more difficult 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 properties article. 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 the inequality by 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.

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.

These points break the axis into intervals.

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.

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).

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: .

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:

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:

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.

How to solve inequalities using the interval method (algorithm with examples)

Example . (task from the OGE) Solve the inequality by the interval method \((x-7)^2< \sqrt{11}(x-7)\)
Solution:

Answer : \((7;7+\sqrt(11))\)

Example . Solve the inequality by the interval method \(≥0\)
Solution:

\(\frac((4-x)^3 (x+6)(6-x)^4)((x+7,5))\)\(≥0\)		Here, at first glance, everything seems normal, and inequality is initially reduced to the desired form. But this is not so - after all, in the first and third brackets of the numerator, x is with a minus sign. We transform the brackets, taking into account the fact that the fourth degree is even (that is, it will remove the minus sign), and the third is odd (that is, it will not remove it). \((4-x)^3=(-x+4)^3=(-(x-4))^3=-(x-4)^3\) \((6-x)^4=(-x+6)^4=(-(x-6))^4=(x-6)^4\) Like this. Now we return the brackets “in place” already converted.
\(\frac(-(x-4)^3 (x+6)(x-6)^4)((x+7,5))\)\(≥0\)		Now all the parentheses look as they should (first comes the unsigned suit, and only then the number). But there was a minus before the numerator. We remove it by multiplying the inequality by \(-1\), not forgetting to reverse the comparison sign
\(\frac((x-4)^3 (x+6)(x-6)^4)((x+7,5))\)\(≤0\)		Ready. Now the inequality looks right. You can use the interval method.
\(x=4;\) \(x=-6;\) \(x=6;\) \(x=-7.5\)		Let's place points on the axis, signs and paint over the necessary gaps.
		In the interval from \(4\) to \(6\), the sign does not need to be changed, because the bracket \((x-6)\) is to an even degree (see paragraph 4 of the algorithm). The flag will be a reminder that the six is also a solution to inequality. Let's write down the answer.

Answer : \((-∞;7,5]∪[-6;4]∪\left\(6\right\)\)

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

\(x^2 (-x^2-64)≤64(-x^2-64)\)		Left and right are the same - this is clearly not accidental. The first desire is to divide by \(-x^2-64\), but this is a mistake, because there is a chance of losing the root. Instead, move \(64(-x^2-64)\) to left side
\(x^2 (-x^2-64)-64(-x^2-64)≤0\)
\((-x^2-64)(x^2-64)≤0\)		Take out the minus in the first bracket and factor the second
\(-(x^2+64)(x-8)(x+8)≤0\)		Note that \(x^2\) is either zero or greater than zero. This means that \(x^2+64\) is uniquely positive for any value of x, that is, this expression does not affect the sign of the left side in any way. Therefore, we can safely divide both parts of the inequality by this expression. Let's also divide the inequality by \(-1\) to get rid of the minus.
\((x-8)(x+8)≥0\)		Now you can apply the interval method
\(x=8;\) \(x=-8\)		Let's write down the answer

Answer : \((-∞;-8]∪}

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Solution of inequalities by the method of intervals. Solving quadratic inequalities by the interval method

Algorithm

Scientific basis of the gap method

Finding zeros of the numerator and denominator

How to solve inequalities using the interval method (algorithm with examples)

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