Solution of rational inequalities by the method of intervals.
The interval method is considered to be universal for solving inequalities. This method is sometimes also referred to as the spacing method. We can apply it 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 method of spans and consider algorithms for solving inequalities using it. Let us touch upon the theoretical aspects on which the application of the method is based.
We pay special attention to the nuances of the topic that are usually not covered in the school curriculum. For example, consider the rules for placing signs on intervals and the method of intervals itself in general form without linking it to rational inequalities.
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Algorithm
Who remembers getting to know the method of spacing in a school algebra course? It usually starts with solving inequalities like f (x)< 0 (знак неравенства может быть использован любой другой, например, ≤ , >or ≥). Here f (x) can be a polynomial or a polynomial ratio. The polynomial, in turn, can be represented as:
- the product of linear binomials with a coefficient of 1 at the variable x;
- product of square trinomials with leading coefficient 1 and negative discriminant of their roots.
Here are some examples of such inequalities:
(x + 3) (x 2 - x + 1) (x + 2) 3 ≥ 0,
(x - 2) (x + 5) x + 3> 0,
(x - 5) (x + 5) ≤ 0,
(x 2 + 2 x + 7) (x - 1) 2 (x 2 - 7) 5 (x - 1) (x - 3) 7 ≤ 0.
Let us write an algorithm for solving inequalities of the form, as we have given in the examples, by the method of intervals:
- we find the zeros of the numerator and denominator, for this we equate the numerator and denominator of the expression on the left side of the inequality to zero and solve the resulting equations;
- we determine the points that correspond to the found zeros and mark them with dashes on the coordinate axis;
- define the signs of the expression f (x) from the left side of the inequality to be solved at each interval and put them on the graph;
- we apply shading over the right sites graphics, guided by the following rule: if the inequality has signs< или ≤ изображается, штрихуются «минусовые» промежутки, если же мы работаем с неравенством, имеющим знаки >or ≥, then we select by hatching the areas marked with the “+” sign.
The cutter that we will be working with may have a schematic view. Excessive detail can overwhelm the drawing and complicate the solution. We will be of little interest to the scale. It will be enough to adhere to the correct location of the points as the values of their coordinates grow.
When working with strict inequalities, we will use the notation of a point in the form of a circle with an open (empty) center. In the case of nonstrict inequalities, the points that correspond to the zeros of the denominator, we will represent 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 a number set, which is actually a solution to a given inequality.
Scientific basis of the method of spacing
The approach underlying the method of intervals 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 characteristic of the number rays (- ∞, a) and (a, + ∞).
The given property of the function is confirmed by the Bolzano-Cauchy theorem, which is given in many textbooks to prepare for the entrance examinations.
It is also possible to justify the constancy of the sign on the intervals based on the properties of numerical inequalities. For example, take the inequality x - 5 x + 1> 0. If we find the zeros of the numerator and denominator and plot them on the number line, we get a series of intervals: (− ∞ , − 1) , (- 1, 5) and (5, + ∞).
We take any of the intervals and show on it that the expression from the left-hand side of the inequality will have a constant sign throughout the entire interval. Let it be the interval (- ∞, - 1). Take any number t from this interval. It will satisfy the conditions t< − 1 , и так как − 1 < 5 , то по свойству транзитивности, оно же будет удовлетворять и неравенству t < 5 .
Using both the 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 say 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 out of between (− ∞ , − 1) ... All this allows us to assert that in the interval taken as an example, the expression has a constant sign. In our case, this is the "+" sign.
Finding the zeros of the numerator and denominator
The algorithm for finding zeros is simple: we equate 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 will limit ourselves to just looking at an example.
Consider the fraction x (x - 0, 6) x 7 (x 2 + 2 x + 7) 2 (x + 5) 3. In order to find the zeros of the numerator and denominator, we equate them to zero in order to obtain and solve the equations: x (x - 0, 6) = 0 and x 7 (x 2 + 2 x + 7) 2 (x + 5) 3 = 0.
In the first case, we can go to the set of two equations x = 0 and x - 0, 6 = 0, which gives us two roots 0 and 0, 6. These are the zeros of the numerator.
The second equation is equivalent to a set of three equations x 7 = 0, (x 2 + 2 x + 7) 2 = 0, (x + 5) 3 = 0. We carry out a number of transformations and get x = 0, x 2 + 2 x + 7 = 0, x + 5 = 0. The root of the first equation is 0, the second equation has no roots, since it has a negative discriminant, the root of the third equation is 5. These are the zeros of the denominator.
0 in this case is both numerator zero and denominator zero.
In the general case, when there is a fraction on the left side of an inequality that is not necessarily rational, the numerator and denominator are likewise equated to zero to obtain equations. Solving the equations allows you to find the zeros of the numerator and denominator.
Determining the sign of the spacing is easy. To do this, you can find the value of the expression on the left side of the inequality for any arbitrary point from the given interval. The resulting sign of the expression value at an arbitrarily selected point in 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 on the left side of the inequality has no zeros in the numerator. The zero of the denominator will be the number - 3. We get two 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 gap (− ∞ , − 3) let's take - 4. At x = - 4 we have (- 4) 2 - (- 4) + 4 (- 4) + 3 = - 24. We received a negative value, which means the entire interval will be with a "-" sign.
For the gap (− 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. Received a positive value, which means that the entire gap will have a "+" sign.
There is another way to define the signs. To do this, we can find a sign at one of the intervals and keep it or change it when crossing zero. In order to do everything correctly, it is necessary to follow the rule: when crossing the 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 the zero of the numerator and the denominator, then the sign can be reversed 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 some examples.
Take the inequality (x - 2) (x - 3) 3 (x - 4) 2 (x - 1) 4 (x - 3) 5 (x - 4) ≥ 0 and solve it by the method of intervals. To do this, we need to find the zeros of the numerator and denominator and mark them on the coordinate line. Numerator zeros are dots 2 , 3 , 4 , the denominator of the point 1 , 3 , 4 . Let's mark them on the coordinate axis with dashes.
We mark the zeros of the denominator with empty dots.
Since we are dealing with a nonstrict inequality, we replace the remaining lines with ordinary dots.
Now let's place the points at the intervals. The rightmost span (4, + ∞) will be a + sign.
Moving from right to left, we will put down the signs of the remaining intervals. We pass through the point with coordinate 4. This is both the numerator and denominator zero. Together, these zeros give the expressions (x - 4) 2 and x - 4... Add their powers 2 + 1 = 3 and get an odd number. This means that the sign in the transition in this case is reversed. The interval (3, 4) will have a minus sign.
We pass to the interval (2, 3) through the point with coordinate 3. This is also the zero of both the numerator and the denominator. We got it thanks to two expressions (x - 3) 3 and (x - 3) 5, the sum of the powers of which is 3 + 5 = 8. Getting an even number allows us to leave the spacing sign unchanged.
Point with coordinate 2 is the zero of the numerator. The degree of expression x - 2 is 1 (odd). This means that when passing through this point, the sign must be changed to the opposite.
We are left with the last interval (- ∞, 1). Point with coordinate 1 is the zero of the denominator. It was derived from the expression (x - 1) 4, with an even degree 4 ... Consequently, the sign remains the same. The final drawing will look like this:
The use of the spacing method is especially effective in cases where evaluating the value of an expression involves a lot 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 in the interval 3 - 3 4, 3 - 2 4.
Now let's start applying the acquired knowledge and skills in practice.
Example 1
Solve the inequality (x - 1) (x + 5) 2 (x - 7) (x - 1) 3 ≤ 0.
Solution
It is advisable to 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 nonstrict inequality, so we mark the zeros of the denominator with empty dots, and the zero of the numerator - 5 with the usual filled point.
Let's sign the spaces using the rules for changing the sign when crossing zero. Let's start with the rightmost interval, for which we calculate the value of the expression from the left-hand side of the inequality at a point arbitrarily taken from the interval. We get the "+" sign. We pass sequentially through all points on the coordinate line, placing signs, and we get:
We are working with a nonstrict inequality that has the sign ≤. This means that we need to mark with hatching the gaps marked with the "-" sign.
Answer: (- ∞ , 1) ∪ (1 , 7) .
The solution of rational inequalities in most cases requires their preliminary transformation to the desired form. Only then does it become possible to use the interval method. Algorithms for carrying out such transformations are considered in the material "Solving rational inequalities".
Consider an example of transforming square trinomials in inequality notation.
Example 2
Find the solution to the inequality (x 2 + 3 x + 3) (x + 3) x 2 + 2 x - 8> 0.
Solution
Let's see if the discriminants of square trinomials in inequality are really negative. This will allow us to determine whether the form of this inequality allows us to apply the method of intervals for the solution.
We calculate the discriminant for the trinomial x 2 + 3 x + 3: D = 3 2 - 4 1 3 = - 3< 0 . Now we calculate the discriminant for the trinomial x 2 + 2 x - 8: D ’= 1 2 - 1 · (- 8) = 9> 0. As you can see, inequality requires a prior transformation. For this, we represent the trinomial x 2 + 2 x - 8 as (x + 4) (x - 2), and then we apply the method of intervals to solve the inequality (x 2 + 3 x + 3) (x + 3) (x + 4) (x - 2)> 0.
Answer: (- 4 , - 3) ∪ (2 , + ∞) .
The generalized spacing 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 specific algorithm. In this case, the algorithm for solving inequalities by the generalized method of intervals will slightly differ from what we analyzed earlier:
- find the domain of definition of the function f and the zeros of this function;
- mark the boundary points on the coordinate axis;
- we put the zeros of the function on the numeric line;
- we determine the signs of the intervals;
- we apply shading;
- write down the answer.
On the number line, it is necessary to mark, among other things, individual points of the domain of definition. For example, the domain of a function is the set (- 5, 1] ∪ (3) ∪ [4, 7) ∪ (10) . This means that we need to mark points with coordinates - 5, 1, 3, 4 , 7 and 10 ... Points − 5 and 7 will be depicted as empty, the rest can be highlighted with a colored pencil in order to distinguish them from the zeros of the function.
The zeros of the function in the case of non-strict inequalities are plotted by ordinary (filled) points, strict ones - by empty points. If the zeros coincide with the boundary points or individual points of the definition area, then they can be repainted in black, making them empty or filled, depending on the type of inequality.
The response record is a numerical set that 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 the inequality contains signs< 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 method of intervals.
Let's 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 the 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:
Let us mark the zero of the function with an empty black dot, since we are working with strict inequality.
We determine the signs at 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 f:
f (16) = 16 2 + 2 16 - 24 - 3 4 16 - 3 16 - 7 = 264 - 15 9> 0 f (8) = 8 2 + 2 8 - 24 - 3 4 8 - 3 8 - 7 = 56 - 9< 0 f (6) = 6 2 + 2 · 6 - 24 - 3 4 · 6 - 3 6 - 7 = 24 - 15 2 - 1 = = 15 - 2 · 24 2 = 225 - 96 2 >0 f (- 8) = - 8 2 + 2 (- 8) - 24 - 3 4 (- 8) - 3 - 8 - 7 = 24 + 3 - 15< 0
We place the just defined signs, and apply shading over the gaps with a minus sign:
The answer will be the union of two intervals with a “-” sign: (- ∞, - 6] ∪ (7, 12).
In response, we included a point with a coordinate of 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 our answer, just as we did not include the entire interval [4, 7). At this point, just like on the entire indicated interval, the value of the function is positive, which does not satisfy the inequality to be solved.
Let's write it down again for a clearer understanding: colored dots must be included in the answer in the following cases:
- these points are part of the hatched gap,
- these points are separate points of the domain of definition of the function, the values of the function in which satisfy the inequality to be solved.
Answer: (− ∞ , − 6 ] ∪ (7 , 12) .
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How to solve inequalities using the interval method (algorithm with examples)
Example
. (task from the OGE) Solve the inequality using the interval method \ ((x-7) ^ 2< \sqrt{11}(x-7)\)
Solution:
Answer : \ ((7; 7+ \ sqrt (11)) \)
Example
... Solve the inequality using the interval method \ (≥0 \)
Solution:
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\ (\ 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 parentheses of the numerator, x is with a minus sign. We transform the parentheses, taking into account the fact that the fourth degree is even (that is, it removes the minus sign), and the third is odd (that is, it does not remove it). |
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\ (\ frac (- (x-4) ^ 3 (x + 6) (x-6) ^ 4) ((x + 7,5)) \)\(≥0\) |
Now all the brackets look as they should (the first is the unsigned claim, and only then the number). But a minus appeared in front of the numerator. We remove it by multiplying the inequality by \ (- 1 \), not forgetting to reverse the comparison sign |
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\ (\ frac ((x-4) ^ 3 (x + 6) (x-6) ^ 4) ((x + 7,5)) \)\(≤0\) |
Ready. Now the inequality looks right. You can apply the method of intervals. |
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\ (x = 4; \) \ (x = -6; \) \ (x = 6; \) \ (x = -7,5 \) |
Let's arrange points on the axis, signs and fill in the necessary intervals. |
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In the interval from \ (4 \) to \ (6 \), the sign does not need to be changed, because the bracket \ ((x-6) \) is in an even power (see point 4 of the algorithm). The flag will be a reminder that the six is also a solution to inequality. |
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:
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\ (x ^ 2 (-x ^ 2-64) ≤64 (-x ^ 2-64) \) |
On the left and on the right there are identical ones - this is clearly no coincidence. The first desire is to divide by \ (- x ^ 2-64 \), but this is a mistake, since there is a chance of losing the root. Move \ (64 (-x ^ 2-64) \) to the left instead |
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\ (x ^ 2 (-x ^ 2-64) -64 (-x ^ 2-64) ≤0 \) |
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\ ((- x ^ 2-64) (x ^ 2-64) ≤0 \) |
Move the minus in the first bracket and factor the second |
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\ (- (x ^ 2 + 64) (x-8) (x + 8) ≤0 \) |
Note: \ (x ^ 2 \) is either zero or greater than zero. This means that \ (x ^ 2 + 64 \) is definitely positive for any x value, that is, this expression does not affect the sign of the left side in any way. Therefore, you can safely divide both sides of the inequality by this expression. |
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\ ((x-8) (x + 8) ≥0 \) |
Now you can apply the spacing method |
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\ (x = 8; \) \ (x = -8 \) |
Let's write down the answer |
Answer : \((-∞;-8]∪}