First level

Interval method. Exhaustive guide (2019)

This method you just need to understand and know it as your five fingers! If only because it is used to solve rational inequalities and because, knowing this method as it should, solve these inequalities surprisingly simply. A little later, I will reveal a couple of secrets, how to save time on solving these inequalities. Well, intrigued? Then we went!

The essence of the method in the decomposition of the inequality of multipliers (replace the topic) and the definition of the OTZ and the sign of the factory, now I will explain everything. Take the most simple example :.

The area of \u200b\u200bpermissible values \u200b\u200b() does not need to write here, since there are no divisions on the variable, and the radicals (roots) are not observed here. The factors here are all so decomposed for us. But do not relax, it's all to remind the basics and understand the essence!

Suppose you do not know the method of intervals, how would you decide this inequality? Come on logically and draw on what you already know. First, the left side will be greater than zero if both expressions in brackets are either more zero, or less zero, because Plus on Plus gives "plus" and "minus" to "minus" gives "plus", right? And if signs in expressions in brackets are different, then in the end the left part will be less than zero. And what do we need to know the meanings in which expressions in brackets will be negative or positive?

We need to solve the equation, it is exactly the same as inequality, only instead of a sign there will be a sign, the roots of this equation will allow you to determine those border values, during the retreat from which multipliers will be larger or less than zero.

And now the intervals themselves. What is the interval? This is a numerical straight line, that is, all possible numbers concluded between two some numbers are the ends of the interval. These intervals are not so easy to submit, so the intervals are taken to draw, now scientific.

We draw the axis, it contains the entire numerical series from and to. On the axis, the points are applied, the most so-called zeros of the functions, the values \u200b\u200bin which the expression is zero. These points "roll out" which means that they do not apply to the number of those values \u200b\u200bin which the inequality is true. In this case, they roll out. A sign in inequality and not, that is, strictly larger and no more or equal.

I want to say that it is not necessary to note zero, he is without circles here, and so, for understanding and orientation along the axis. Okay, the axis was painted, the points (more precisely the mug) put, further what, how would it help me in solving? - You ask you. Now just take the value for the ICA from the intervals in order and submese them into your inequality and see which sign will be as a result of multiplication.

In short, just take, for example, we substitute it here, it will turn out, and it means on the whole interval (throughout the interval) from to which we took, inequality will be fair. In other words, if X is from before, then the inequality is true.

I do the same with the interval from before, we take or, for example, we substitute in, we define the sign, the mark will be "minus". And we do the same with Postgraduate, the third interval from to, where the sign will be "plus". Such a lot of text came out, but a little clarity is true?

Looking at once again for inequality.

Now everything is also applied to the same axis and signs that will result in the result. The broken line, in my example, we indicate the positive and negative sections of the axis.

Look at the inequality - on the drawing, again for inequality - and again to the drawingIs something clear? Try now to say at what intervals of the ICA, the inequality will be true. That's right, from to the inequality will be true and from before, and at the interval from to the inequality of zero, and this gap is little interests, because we have a sign in inequality.

Well, since you figured out this, then it's small - to write out the answer! In response, we write those gaps at which the left side is more than zero, which is read like an X-line belongs from the minus of infinity to minus one and two to plus infinity. It is worth clarifying that parentheses mean that the values \u200b\u200blimited by the interval are not solutions of inequality, that is, they are not included in response, but only suggest that before, for example, but not a solution.

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

What do you think, what should be done before the point on the axis to apply? Yeah, the factors decompose:

We draw the intervals and set the signs, notice the points from us to be frozen, because the sign is strictly less than zero:

It's time to reveal to you one secret, which I promised at the beginning of this topic! And what if I tell you that you can not substitute the values \u200b\u200bfrom each interval to determine the sign, but you can define a sign in one of the intervals, and in the others just alternate signs!

Thus, we saved a little time on an affixation of characters - I think this time won on the exam does not prevent!

We write the answer:

Now consider an example of a fractional rational inequality - inequality, both parts of which are rational expressions (see).

What can you say about this inequality? And you look at him like fractional rational equationWhat are we doing first? Immediately we see that there is no roots, it means definitely rational, but immediately fraction, and even with an unknown in the denominator!

True, OTZ need!

So, further went, here all the factors except one have a variable of the first degree, but there is a multiplier where X is the second degree. Usually, the sign changed after the transition through one of the points in which the left part of the inequality takes the zero value, for which we determined what the X is equal to EX in each multiplier. And here, so it is always positive, because Any number in square\u003e zero and positive terms.

What do you think will affect the meaning of inequality? Right - will not affect! We can safely divide on both parts of inequality and thereby remove this multiplier so that the eyes do not call.

it was time to draw the intervals to draw, for this it is necessary to define those border values, during the retreat from which multipliers will be larger and less than zero. But pay attention that here the sign means the point in which the left part of the inequality takes the zero value, we will not pump out, it is among the solutions, such a point we have one, is the point where X is equal to one. And the point where the denominator is negative to the core? - Of course not!

The denominator should not be zero, so the interval will look like this:

For this scheme, you can easily write an answer, I just say that now you have at your disposal there is a new bracket type - square! Here is such a bracket [ It says that the value is included in the solutions interval, i.e. It is part of the answer, this bracket corresponds to the painted (not painted) point on the axis.

Here, - did you get the same answer?

We decompose on the factors and transfer everything in one direction, we only need to be left to the right, to compare with it:

I pay your attention that in the last transformation, in order to get in the numerator as in the denominator, I multiply both parts of inequality. Remember that when multiplying both parts of inequality on, the sign of inequality changes to the opposite !!!

We write ...

Otherwise, the denominator will turn to zero, and on zero, as you remember, it is impossible to share!

Agree, in the resulting inequality, it is waved to cut in a numerator and denominator! It is impossible to do this, you can lose some of the decisions or ...

Now try to apply points on the axis. I note only that when applying points it is necessary to pay attention to the fact that the point with a value that proceeds from the sign seemingly should be applied to the axis like painted, it will not be painted, it will be inquiring! Why do you ask? And you even remember, you are not going to share it for zero?

Remember, OWN is over all! If all the inequality and signs of equality say one thing, and OTZ is another, trust OST, Great and Mighty! Well, you built intervals, I'm sure you took advantage of my hint about the alternation and you did it like this (see the drawing below) And now you are smoking, and do not repeat this error more! What mistake? - You ask you.

The fact is that in this inequality the multiplier was repeated twice (remember how you still rushed it?). So, if a multiplier is repeated in inequality an even number of times, then when switching through the point on the axis, which draws this multiplier to zero (in this case, the point) will not change the sign, if an odd, then the sign changes!

It will be faithful to the following axis with intervals and signs:

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

Answer:

I will say that there are situations where there are the roots of inequalities that are not included in any interval, in response they are recorded in curly brackets, like this, for example :. You can read more about such situations in the Middle Level article.

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

1. We carry everything into the left part, we leave only zero on the right;
2. We find ...
3. We apply on the axis all the roots of inequality;
4. We take arbitrary from one of the gaps and determine the sign in the interval to which the root belongs, alternate signs, paying attention to the roots, repeating in inequality several times, from parity or count the amount of their repetition, changes the sign when passing through them or not;
5. In response, we write intervals, observing the scrubber and not paint points (see OTZ), putting the necessary types of brackets between them.

Well, finally, our favorite heading, "do it yourself"!

Examples:

Answers:

Interval method. AVERAGE LEVEL

Linear function

Linear is called the function of the form. Consider the function for example. It is positive at and negative when. Point is zero function (). Let's show the signs of this function on the numeric axis:

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

It can be seen that the functions of the function correspond to the position of the function of the function: if the schedule is above the axis, the sign "", if below - "".

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

• We find zero functions;
• We note it on the numeric axis;
• Determine the sign of the function on different sides of zero.

Quadratic function

I hope you remember how square inequalities are solved? If not, read the topic. Let's remind general form quadratic function: .

Now let's remember which signs receive a quadratic function. Its graph - Parabola, and the function takes the sign "" with such in which Parabola is above the axis, and "" - if Parabola is below the axis:

If the function has zeros (values \u200b\u200bin which), Parabola crosses the axis at two points - the roots of the corresponding square equation. Thus, the axis is divided into three intervals, and the signs of the function alternately change when moving through each root.

Is it possible to somehow define signs without drawing every time a parabola?

Recall that the square three decrease can be decomposed on the factors:

For example: .

Note roots on the axis:

We remember that the function sign can only change when moving through the root. We use this fact: for each of the three intervals to which the axis is divided with roots, it is enough to determine the function of the function only in one arbitrarily selected point: at the other points of the interval the sign will be the same.

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

Well, when (submits, for example,) both brackets are negative, it means that the work is positive:

That's what it is interval method: Knowing the signs of the factors at each interval, we define the sign of all the work.

Consider also cases when there is no zeros of the function, or it is only one.

If there are no, then there are no roots. So, there will be no "transition through the root". So, the function on the entire numeric axis takes only one sign. It is easy to determine, substituting into the function.

If the root is only one, the parabol is touched by the axis, so the function sign does not change when moving through the root. What rule will come up for such situations?

If you decompose such a function on multipliers, two identical multipliers will turn out:

And any expression in the square is nonnegative! Therefore, the function of the function does not change. In such cases, we will allocate the root, when moving through which the sign does not change, circled with a square:

Such root will be called multiple.

Interval method in inequalities

Now any square inequality can be solved without drawing a parabola. It is enough just to place the signs of the quadratic function on the axis, and select the intervals depending on the sign of inequality. For example:

Mind roots on the axis and lay signs:

We need part of the axis with the sign ""; Since the inequality of the unrest, the roots themselves are also included in the solution:

Now consider rational inequality - inequality, both parts of which are rational expressions (see).

Example:

All factors except one - - here "linear", that is, contain a variable only in the first degree. Such linear multipliers are needed to apply the interval method - a sign when moving through their roots changes. But the multiplier does not have roots at all. This means that it is always positive (check it yourself), and therefore does not affect the sign of all inequality. It means that it can be divided by the left and right-hand side of the inequality, and thus get rid of it:

Now everything is the same as it was with square inequalities: we determine which points each of the multipliers turn to zero, mark these points on the axis and arrange signs. I pay attention very important fact:

Answer:. Example:.

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

In a numerator and denominator, the same multiplier, but not in a hurry to cut it! After all, then we can forget to buy this point. It is better to note this root as a multiple, that is, when moving through it, the sign will not change:

Answer:.

And one more very demonstration example:

Again, we do not reduce the same multipliers of the numerator and the denominator, since if we reduce, we will have to specifically memorize that you need to buy a point.

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

In the case of an even number, we do the same as before: we supply the point by the square and do not change the sign when moving through the root. But in the case of an odd amount, this rule is not executed: the sign will still be changed during the transition through the root. Therefore, with such a root, we do not additionally do nothing, as if it is not a multiple. The above rules relate to all even and odd degrees.

What we write in the answer?

If there is a violation of the alternation of signs, it is necessary to be very attentive, because with incomprehensible inequality in response all painted points. But some of Nah often stand a mansion, that is, not included in the painted area. In this case, we add them to the answer as insulated points (in curly brackets):

Examples (Solving yourself):

Answers:

1. If among the multipliers is simply the root, because it can be represented as.
   .

Interval method. Briefly about the main thing

The interval method is used to solve rational inequalities. It lies in determining the sign of the work on the signs of factors at various intervals.

Algorithm for solving rational inequalities by intervals.

• We carry everything into the left part, we leave only zero on the right;
• We find ...
• We apply on the axis all the roots of inequality;
• We take arbitrary from one of the gaps and determine the sign in the interval to which the root belongs, alternate signs, paying attention to the roots, repeating in inequality several times, from parity or count the amount of their repetition, changes the sign when passing through them or not;
• In response, we write intervals, observing the scrubber and not paint points (see OTZ), putting the necessary types of brackets between them.

Well, the topic is finished. If you read these lines, then you are very cool.

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Fill a hand by solving tasks on this topic.

You will not ask the theory on the exam.

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And if you did not solve them (a lot!), You definitely be a foolishly mistaken or just do not have time.

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

Find where you want a collection, mandatory with solutions, detailed analysis And decide, decide, decide!

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Find the task and decide!

In this lesson, we will continue to solve rational inequalities by intervals for more complex inequalities. Consider the solution of fractional linear and fractional-quadratic inequalities and related tasks.

Now we return to inequality

Consider some associated tasks.

Find the smallest solution of inequality.

Find the number of natural solutions inequality

Find the length of the intervals that make up many solutions of inequality.

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1. Mordkovich A.G. and others. Algebra 9 CL.: Task for students of general education institutions / A. Mordkovich, T. N. Mishoustina, etc. - 4th ed. - M.: Mnemozina, 2002.-143 s.: Il. №№ 28 (B, B); 29 (b, c); 35 (a, b); 37 (b, c); 38 (a).

How to solve inequalities by intervals (algorithm with examples)

Example . (Task from OGE) Solve the inequality of intervals \\ ((x-7) ^ 2< \sqrt{11}(x-7)\)
Decision:

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

Example . Decide the inequality of the interval method \\ (≥0 \\)
Decision:

\\ (\\ FRAC ((4-x) ^ 3 (x + 6) (6-x) ^ 4) ((X + 7.5)) \\)\(≥0\)		Here at first glance everything seems normal, and the inequality initially shown to the right mind. But this is not the case - after all, in the first and third bracket, the Numerator of the Iks stands with a minus sign. We transform brackets, taking into account that the fourth degree is even (that is, the minus sign will remove), and the third is odd (that is, will not remove). \\ ((4-x) ^ 3 \u003d (- x + 4) ^ 3 \u003d (- (x-4)) ^ 3 \u003d - (x-4) ^ 3 \\) \\ ((6-x) ^ 4 \u003d (- x + 6) ^ 4 \u003d (- (x-6)) ^ 4 \u003d (x-6) ^ 4 \\) Like this. Now we return the brackets "in place" already transformed.
\\ (\\ FRAC (- (x-4) ^ 3 (x + 6) (x-6) ^ 4) ((x + 7.5)) \\)\(≥0\)		Now all the brackets look like it (the first is the lawsuit without a sign and only then the number). But before the numerator appeared minus. Remove it, multiplying inequality on \\ (- 1 \\), not forgetting to turn the comparison sign
\\ (\\ FRAC ((X-4) ^ 3 (X + 6) (X-6) ^ 4) ((x + 7.5)) \\)\(≤0\)		Ready. Now the inequality looks like. You can use the interval method.
\\ (x \u003d 4; \\) \\ (x \u003d -6; \\) \\ (x \u003d 6; \\) \\ (x \u003d -7.5 \\)		We put the points on the axis, signs and custody the necessary intervals.
		In the interval from \\ (4 \\) to \\ (6 \\), the sign should not be changed, because the bracket \\ ((x-6) \\) to the even degree (see paragraph 4 of the algorithm). The checkbox will be a reminder that the six is \u200b\u200balso a decision of inequality. We write the answer.

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

Example. (Task from OGE) Solve the inequality by the method of intervals \\ (x ^ 2 (-x ^ 2-64) ≤64 (-x ^ 2-64) \\)
Decision:

\\ (x ^ 2 (-x ^ 2-64) ≤64 (-x ^ 2-64) \\)		On the left and right there are the same - it is clearly not by chance. The first desire is to divide on \\ (- x ^ 2-64 \\), but this is a mistake, because There is a chance to lose root. Instead, we transfer \\ (64 (-x ^ 2-64) \\) on the left side
\\ (x ^ 2 (-x ^ 2-64) -64 (-x ^ 2-64) ≤0 \\)
\\ ((- x ^ 2-64) (x ^ 2-64) ≤0 \\)		I will resert a minus in the first bracket and spread the second to multipliers
\\ (- (x ^ 2 + 64) (x-8) (x + 8) ≤0 \\)		Note: \\ (x ^ 2 \\) either equal to zero or more zero. So, \\ (x ^ 2 + 64 \\) - uniquely positively at any value of the ICA, that is, this expression does not affect the sign of the left side. Therefore, you can safely share both parts of inequality for this expression. We divide the inequality just on \\ (- 1 \\) to get rid of the minus.
\\ ((x-8) (x + 8) ≥0 \\)		Now you can apply the interval method
\\ (x \u003d 8; \\) \\ (x \u003d -8 \\)		We write the answer

Answer : \((-∞;-8]∪}