Continuity of function. Break points.

A goby walks, sways, sighs on the go:
- Oh, the board ends, now I'm going to fall!

In this lesson, we will analyze the concept of continuity of a function, the classification of break points and a common practical problem. function continuity studies... From the very name of the topic, many intuitively guess what will be discussed, and think that the material is quite simple. This is true. But it is simple tasks that are most often punished for neglect and a superficial approach to their solution. Therefore, I recommend that you study the article very carefully and catch all the subtleties and techniques.

What do you need to know and be able to do? Not very much. For a high-quality assimilation of a lesson, you need to understand what it is function limit... Readers with low level preparation is enough to comprehend the article Limits of functions. Examples of solutions and see the geometric meaning of the limit in the manual Graphs and properties of elementary functions... It is also advisable to familiarize yourself with geometric transformations of graphs, since practice in most cases involves building a drawing. Prospects are optimistic for everyone, and even a full kettle will be able to cope with the task on its own in the next hour or two!

Continuity of function. Break points and their classification

Continuity of a function

Consider some function that is continuous on the whole number line:

Or, more laconically speaking, our function is continuous on (the set of real numbers).

What is the “philistine” criterion of continuity? Obviously, the graph of a continuous function can be drawn without lifting the pencil from the paper.

In this case, two simple concepts should be clearly distinguished: function domain and continuity of function... In general they are not the same... For example:

This function is defined on the whole number line, that is, for of each the meaning of "x" there is a meaning of "game". In particular, if, then. Note that the other point is punctured, because by the definition of the function, the value of the argument must match the only thing function value. Thus, domain our function:.

but this function is not continuous on! It's clear that at the point she endures break... The term is also quite intelligible and descriptive, indeed, the pencil here will have to be torn off the paper anyway. A little later, we will look at the classification of break points.

Continuity of a function at a point and on an interval

In one way or another math problem we can talk about the continuity of a function at a point, the continuity of a function on an interval, a half-interval, or the continuity of a function on a segment. That is, there is no "just continuity"- the function can be continuous WHERE-THAT. And the fundamental building block of everything else is continuity of function at the point .

The theory of mathematical analysis gives a definition of the continuity of a function at a point using the "delta" and "epsilon" neighborhoods, but in practice there is another definition in use, which we will pay the closest attention to.

Let's remember first one-sided limits who burst into our lives in the first lesson about function graphs... Consider an everyday situation:

If you approach the axis along the point left(red arrow), then the corresponding values ​​of the "players" will go along the axis to the point (crimson arrow). Mathematically, this fact is fixed using left-hand limit:

Pay attention to the entry (it reads "x tends to ka on the left"). "Additive" "minus zero" symbolizes , in fact, this means that we are approaching the number from the left side.

Similarly, if you approach the point "ka" on right(blue arrow), then the "gamers" will come to the same value, but already along the green arrow, and right-hand limit will be formalized as follows:

"Additive" symbolizes , and the entry reads like this: "x tends to ka on the right."

If the one-sided limits are finite and equal(as in our case): , then we will say that there is a GENERAL limit. It's simple, the general limit is our "usual" function limit equal to a finite number.

Note that if the function is not defined for (poke out black point on the graph branch), the above calculations remain valid. As has already been noted many times, in particular, in the article on infinitesimal functions, expressions mean that "x" infinitely close approaches the point, while DOES NOT MATTER whether the function itself is defined at a given point or not. Good example appears in the next paragraph when the function is analyzed.

Definition: a function is continuous at a point if the limit of the function at this point is equal to the value of the function at this point:.

The definition is detailed in the following conditions:

1) The function must be defined at a point, that is, a value must exist.

2) There must be an overall function limit. As noted above, this implies the existence and equality of unilateral limits: .

3) The limit of the function at this point must be equal to the value of the function at this point:.

If violated at least one from three conditions, then the function loses its continuity property at the point.

Continuity of the function on the interval is formulated in a clever and very simple way: a function is continuous on an interval if it is continuous at every point of a given interval.

In particular, many functions are continuous on an infinite interval, that is, on the set of real numbers. This is a linear function, polynomials, exponential, sine, cosine, etc. And in general, any elementary function continuous on its areas of definition, so, for example, the logarithmic function is continuous on the interval. Hope to this moment you have a pretty good idea of ​​what the graphs of the main functions look like. More detailed information their continuity can be found in kind person by the name of Fichtengolts.

With the continuity of a function on a segment and half-intervals, everything is also easy, but it is more appropriate to talk about this in the lesson on finding the minimum and maximum values ​​of the function on the segment, but for now we will not hammer our heads.

Break point classification

The fascinating life of functions is rich in all sorts of special points, and the breaking points are just one of the pages of their biography.

Note : just in case, I will focus on an elementary moment: a breakpoint is always single point- there are no "several break points in a row", that is, there is no such thing as "break interval".

These points, in turn, are divided into two large groups: breaks of the first kind and breaks of the second kind... Each gap type has its own characteristics, which we will look at right now:

Breakpoint of the first kind

If the continuity condition is violated at a point and one-sided limits are finite then it is called break point of the first kind.

Let's start with the most optimistic case. According to the initial idea of ​​the lesson, I wanted to tell the theory "in general view”, But in order to demonstrate the reality of the material, he settled on the option with specific characters.

Sadly, like a photo of newlyweds in front of the Eternal Flame, but the following frame is generally accepted. Let's draw a graph of the function in the drawing:


This function is continuous on the whole number line, except for the point. Indeed, the denominator cannot be is zero... However, in accordance with the meaning of the limit - we can infinitely close to approach "zero" both to the left and to the right, that is, one-sided limits exist and, obviously, coincide:
(Condition No. 2 of continuity is satisfied).

But the function is not defined at the point, therefore, the condition No. 1 of continuity is violated, and the function suffers a discontinuity at this point.

A gap of this kind (with the existing general limit) are called removable gap... Why disposable? Because the function can be redefine at the break point:

Looks strange? Perhaps. But this function does not contradict anything! Now the gap has been closed and everyone is happy:


Let's do a formal check:

2) - there is a general limit;
3)

Thus, all three conditions are satisfied, and the function is continuous at a point by the definition of the continuity of a function at a point.

However, haters of matan can define the function in a bad way, for example :


It is curious that the first two conditions of continuity are fulfilled here:
1) - the function is defined at a given point;
2) - there is a general limit.

But the third milestone is not passed:, that is, the limit of the function at the point not equal the value of this function at this point.

Thus, the function breaks at a point.

The second, sadder case is called break of the first kind with a jump... And sadness is evoked by one-sided limits that are finite and different... An example is shown in the second drawing of the lesson. Such a gap occurs, as a rule, in piecewise-defined functions already mentioned in the article about graph transformations.

Consider a piecewise function and we will execute its drawing. How to build a graph? Very simple. On the half-interval we draw a fragment of a parabola (green), on the interval - a straight line segment (red) and on the half-interval - a straight line ( blue color).

Moreover, due to inequality, the value is determined for quadratic function(green point), and due to inequality, the value is defined for a linear function (blue point):

In the very hard case one should resort to the point-by-point construction of each piece of the graph (see the first lesson about function graphs).

Now we will only be interested in the point. Let's examine it for continuity:

2) Let's calculate the one-sided limits.

On the left we have a red line segment, so the left-side limit is:

On the right is the blue line, and the right-hand limit:

As a result, received finite numbers and they not equal... Since the one-sided limits are finite and different: then our function suffers break of the first kind with a jump.

It is logical that the gap cannot be eliminated - the function really cannot be redefined and "glued together", as in the previous example.

Breakpoints of the second kind

Usually all other rupture cases are slyly classified in this category. I will not list everything, because in practice, in 99% percent of tasks you will encounter endless break- when left-handed or right-handed, and more often, both limits are infinite.

And, of course, the most suggestive picture is the hyperbole at point zero. Here, both one-sided limits are infinite: , therefore, the function suffers a discontinuity of the second kind at a point.

I try to fill my articles with as varied content as possible, so let's take a look at a function graph that we haven't seen yet:

according to the standard scheme:

1) The function is not defined at this point because the denominator vanishes.

Of course, one can immediately conclude that the function has a break at a point, but it would be good to classify the nature of the break, which is often required by condition. For this:

I remind you that the recording means infinitesimal a negative number , and under the entry - infinitesimal positive number.

The one-sided limits are infinite, which means that the function suffers a discontinuity of the 2nd kind at a point. The ordinate axis is vertical asymptote for the schedule.

It is not uncommon for both one-sided limits to exist, but only one of them is infinite, for example:

This is the graph of the function.

Let us examine the point for continuity:

1) The function is not defined at this point.

2) Let's calculate the one-sided limits:

We will talk about the methodology for calculating such one-sided limits in the last two examples of the lecture, although many readers have already seen and guessed everything.

The left-side limit is finite and equal to zero (we “do not go” to the point itself), but the right-side limit is infinite and the orange branch of the graph is infinitely close to its vertical asymptote, given by the equation(black dotted line).

So the function suffers break of the second kind at the point.

As for a break of the 1st kind, at the very point of break, the function can be defined. For example, for a piecewise function feel free to put a black bold point at the origin. On the right is a branch of hyperbole, and the right-side limit is infinite. I think almost everyone has an idea of ​​what this graph looks like.

What everyone was looking forward to:

How to investigate a function for continuity?

The study of the function for continuity at a point is carried out according to the already knurled routine scheme, which consists in checking three conditions of continuity:

Example 1

Explore function

Solution:

1) The only point in which the function is not defined falls under the sight.

2) Let's calculate the one-sided limits:

One-sided limits are finite and equal.

Thus, at a point, the function suffers a removable discontinuity.

What does the graph of this function look like?

I would like to simplify , and it seems to be an ordinary parabola. BUT the original function is not defined at the point, so the following caveat is required:

Let's execute the drawing:

Answer: the function is continuous on the whole number line except for the point at which it suffers a removable discontinuity.

The function can be redefined in a good or bad way, but by condition it is not required.

You say, a contrived example? Not at all. We met dozens of times in practice. Almost all the tasks of the site come from real independent and control works.

Let's get rid of our favorite modules:

Example 2

Explore function for continuity. Determine the nature of the function gaps, if they exist. Execute a blueprint.

Solution: for some reason, students are afraid and do not like functions with a module, although there is nothing complicated about them. We have already touched on such things a little in the lesson. Geometric transformations of graphs... Since the modulus is non-negative, it is expanded as follows: , where "alpha" is some expression. In this case, our function should be signed in a piecewise manner:

But the fractions of both pieces have to be reduced by. Reduction, as in the previous example, will not go without consequences. The original function is undefined at the point because the denominator vanishes. Therefore, the system should additionally indicate a condition, and make the first inequality strict:

Now about a VERY USEFUL solution: before finishing the task on a draft, it is beneficial to make a drawing (regardless of whether it is required by condition or not). This will help, firstly, to immediately see the points of continuity and breakpoints, and, secondly, it will 100% save you from mistakes when finding one-sided limits.

Let's complete the drawing. In accordance with our calculations, to the left of the point it is necessary to draw a fragment of a parabola (blue), and to the right - a piece of a parabola (red), while the function is not defined at the point itself:

If in doubt, take several "x" values, plug them into the function (not forgetting that the module destroys the possible minus sign) and check the graph.

Let us investigate the function for continuity analytically:

1) The function is not defined at a point, so we can immediately say that it is not continuous at it.

2) Establish the nature of the discontinuity, for this we calculate the one-sided limits:

One-sided limits are finite and different, which means that the function suffers a discontinuity of the 1st kind with a jump at a point. Note again that when finding the limits, it doesn't matter whether the function is defined at the break point or not.

Now it remains to transfer the drawing from the draft (it was made as if with the help of research ;-)) and complete the task:

Answer: the function is continuous on the whole number line except for the point at which it suffers a discontinuity of the first kind with a jump.

Sometimes it is required to additionally indicate the gap jump. It is calculated in an elementary way - from the right limit, you need to subtract the left limit:, that is, at the break point, our function jumped 2 units down (as indicated by the minus sign).

Example 3

Explore function for continuity. Determine the nature of the function gaps, if they exist. Make a drawing.

This is an example for independent decision, a sample solution at the end of the tutorial.

Let's move on to the most popular and widespread version of the task, when the function consists of three parts:

Example 4

Examine the function for continuity and graph the function .

Solution: it is obvious that all three parts of the function are continuous on the corresponding intervals, so it remains to check only two points of the "joint" between the pieces. First, let's make a drawing on a draft; I commented out the construction technique in sufficient detail in the first part of the article. The only thing you need to carefully follow our special points: due to inequality, the value belongs to a straight line (green point), and due to inequality, the value belongs to a parabola (red point):


Well, in principle, everything is clear =) It remains to make a decision. For each of the two "butting" points, we standardly check 3 conditions of continuity:

I) Let us investigate the point

1)

One-sided limits are finite and different, which means that the function suffers a discontinuity of the 1st kind with a jump at a point.

We calculate the discontinuity jump as the difference between the right and left limits:
, that is, the chart jumped one unit up.

II) Let us investigate the point

1) - the function is defined at a given point.

2) Find one-sided limits:

- one-sided limits are finite and equal, which means that there is a common limit.

3) - the limit of a function at a point is equal to the value of this function at a given point.

At the final stage, we transfer the drawing to the final copy, after which we put the final chord:

Answer: the function is continuous on the whole number line, except for the point at which it suffers a discontinuity of the first kind with a jump.

Example 5

Examine the function for continuity and plot its graph .

This is an example for an independent solution, a short solution and a rough example of how to design a problem at the end of the lesson.

One might get the impression that at one point the function must necessarily be continuous, and at another there must be a gap. In practice, this is far from always the case. Try not to neglect the remaining examples - there will be several interesting and important chips:

Example 6

The function is given ... Examine the function for continuity at points. Build a graph.

Solution: and again immediately execute the drawing on the draft:

The peculiarity of this graph is that at, the piecewise function is given by the equation of the abscissa axis. This section is drawn here. green, and in a notebook it is usually highlighted in bold with a simple pencil. And, of course, do not forget about our rams: the value belongs to the tangent branch (red dot), and the value belongs to the straight line.

Everything is clear from the drawing - the function is continuous on the entire number line, it remains to draw up a solution, which is brought to complete automatism literally after 3-4 such examples:

I) Let us investigate the point

1) - the function is defined at this point.

2) Let's calculate the one-sided limits:

so there is a general limit.

For every fireman, let me remind you of a trivial fact: the limit of a constant is equal to the constant itself. In this case, the zero limit is zero itself (left-handed limit).

3) - the limit of a function at a point is equal to the value of this function at a given point.

Thus, a function is continuous at a point by the definition of the continuity of a function at a point.

II) Let us investigate the point

1) - the function is defined at this point.

2) Find one-sided limits:

And here - the limit of the unit is equal to the unit itself.

- there is a general limit.

3) - the limit of a function at a point is equal to the value of this function at a given point.

Thus, a function is continuous at a point by the definition of the continuity of a function at a point.

As usual, after research, we transfer our drawing to a clean copy.

Answer: the function is continuous at points.

Please note that in the condition we were not asked about the study of the entire function for continuity, and it is considered good mathematical form to formulate precise and precise the answer to the question posed. By the way, if by condition it is not required to build a schedule, then you have every right not to build it (however, then the teacher can force you to do it).

A small mathematical "tongue twister" for an independent solution:

Example 7

The function is given ... Examine the function for continuity at points. Classify breakpoints, if any. Execute a blueprint.

Try to correctly "pronounce" all the "words" =) And draw the graph more precisely, accuracy, it will not be superfluous everywhere ;-)

As you remember, I recommended to immediately carry out a drawing on a draft, but from time to time I come across such examples where you cannot immediately figure out what the graph looks like. Therefore, in a number of cases, it is beneficial to first find one-sided limits and only then, on the basis of research, depict the branches. In the two final examples, we will also master the technique of calculating some one-sided limits:

Example 8

Examine the function for continuity and plot its schematic graph.

Solution: bad points are obvious: (turns the denominator of the indicator to zero) and (turns to zero the denominator of the whole fraction). It is not clear what the graph of this function looks like, which means that it is better to do some research first.

Determining the breakpoint of a function
End point x 0 called function discontinuity point f (x) if the function is defined on some punctured neighborhood of the point x 0 but is not continuous at this point.

That is, at the point of discontinuity, the function is either undefined or defined, but at least one one-sided limit at this point either does not exist, or is not equal to the value of f (x 0) functions at the point x 0 ... See “Determining Continuity of a Function at a Point”.

Determination of the break point of the 1st kind
The point is called break point of the first kind if is a break point and there are finite one-sided left and right limits:
.

Defining a jump function
The jump Δ function at a point is the difference between the limits on the right and left
.

Determination of the point of a removable discontinuity
The point is called point of removable discontinuity if there is a limit
,
but the function at the point is either undefined or not equal to the limit value:.

Thus, the point of a removable discontinuity is a point of discontinuity of the first kind, at which the jump of the function is equal to zero.

Determination of the break point of the 2nd kind
The break point is called break point of the second kind if it is not a breakpoint of the 1st kind. That is, if at least one one-sided limit does not exist, or at least one one-sided limit at a point is equal to infinity.

Examining functions for continuity

When investigating functions for continuity, we use the following facts.

• Elementary functions and their inverse are continuous on their domain of definition. These include the following features:
  , as well as constant and inverse functions. See "Basic Functions Reference".
• Sum, difference and product continuous, on some set of functions, is continuous, function on this set.
  Private two continuous, on some set of functions, is a continuous function on this set, except for the points at which the denominator of the fraction vanishes. See "Arithmetic Properties of Continuous Functions"
• Complex function is continuous at a point if the function is continuous at the point and the function is continuous at the point. See "Limit and Continuity of a Complex Function"

Examples of

Example 1

A function and two argument values ​​and are specified. It is required: 1) to establish whether the given function is continuous or discontinuous for each of the given values ​​of the argument; 2) in the case of a discontinuity of a function, find its limits at the point of discontinuity on the left and right, establish the type of discontinuity; 3) make a schematic drawing.
.

The given function is complex. It can be seen as a composition of two functions:
,. Then
.

Let's consider a function. It is composed of a function and constants using the arithmetic operations of addition and division. The function is elementary - a power function with an exponent 1 ... It is defined and continuous for all values ​​of the variable. Therefore, the function is defined and continuous for all, except for the points at which the denominator of the fraction vanishes. Equate the denominator to zero and solve the equation:
.
We get a single root.
So, the function is defined and continuous for all but the point.

Let's consider a function. It is an exponential function with a positive base of the degree. It is defined and continuous for all values ​​of the variable.
That's why preset function is defined and continuous for all values ​​of the variable, except for the point.

Thus, at the point, the given function is continuous.

Function graph y = 4 1 / (x + 2).

Consider a point. The function is undefined at this point. Therefore, it is not continuous. Let's establish the kind of gap. To do this, we find one-sided limits.

Using the connection between infinitely large and infinitely small functions, for the limit on the left we have:
at ,
,
,
.

Here we have used the following generally accepted notation:
.
We also used the property of the exponential function with a radix:
.

Similarly, for the limit on the right we have:
at ,
,
,
.

Since one of the one-sided limits is equal to infinity, there is a discontinuity of the second kind at the point.

The function is continuous at a point.
At the point there is a break of the second kind,
.

Example 2

The function is set. Find the breakpoints of the function, if they exist. Specify the kind of gap and function jumps, if any. Make a drawing.
.

Schedule of a given function.

The function is power function with an integer exponent equal to 1 ... This function is also called linear. It is defined and continuous for all values ​​of the variable.

It includes two more functions: and. They are composed of a function and constants using the arithmetic operations of addition and multiplication:
, .
Therefore, they are also continuous for everyone.

Since the functions included in the composition are continuous for all, it can have break points only at the gluing points of its components. These are points and. Let us investigate for continuity at these points. For this, we find one-sided limits.

Consider a point. To find the left limit of a function at this point, we must use the values ​​of this function in any left punctured neighborhood of the point. Let's take the neighborhood. On it. Then the limit is on the left:
.
Here we have used the fact that the function is continuous at a point (just like at any other point). Therefore, its left (as well as right) limit is equal to the value of the function at this point.

Let's find the right limit at the point. To do this, we must use the values ​​of the function in any right punctured neighborhood of this point. Let's take the neighborhood. On it. Then the limit is on the right:
.
We also took advantage of function continuity here.

Since, at the point, the limit on the left is not equal to the limit on the right, then the function in it is not continuous - it is a discontinuity point. Since the one-sided limits are finite, this is a breakpoint of the first kind. Jumping function:
.

Now let's look at the point. We calculate the one-sided limits in the same way:
;
.
Since the function is defined at a point and the left limit is equal to the right one, the function is continuous at this point.

The function has a discontinuity of the first kind at a point. Jump functions in it:. The function is continuous at other points.

Example 3

Determine the points of discontinuity of the function and investigate the nature of these points if
.

We will use the fact that the linear function is defined and continuous for all. The given function is composed of a linear function and constants using the arithmetic operations of addition, subtraction, multiplication and division:
.
Therefore, it is defined and continuous for all, except for the points at which the denominator of the fraction vanishes.

Let's find these points. Equate the denominator to zero and solve the quadratic equation:
;
;
; .
Then
.

We use the formula:
.
With its help, we factor out the numerator:
.

Then the given function will take the form:
(W1) .
It is defined and continuous for all, except for points and. Therefore, the points and are the points of discontinuity of the function.

Divide the numerator and denominator of the fraction in (A1) by:
(P2) .
We can do this operation if. Thus,
at .
That is, the functions and differ only at one point: it is defined at, and at this point it is not defined.

To determine the genus of the breakpoints, we need to find the one-sided limits of the function at the points and. To calculate them, we will use the fact that if the values ​​of the function are changed, or made undefined at a finite number of points, then this will not have any effect on the magnitude or existence of the limit at an arbitrary point (see “Influence of the values ​​of a function at a finite number of points on the value of the limit "). That is, the limits of the function at any points are equal to the limits of the function.

Consider a point. The denominator of the fraction in the function does not vanish at. Therefore, it is defined and continuous at. It follows that there is a limit at and it is equal to the value of the function at this point:
.
Therefore, the point is a point of a removable discontinuity of the first kind.

Consider a point. Using the connection between infinitely small and infinitely large functions, we have:
;
.
Since the limits are infinite, there is a gap of the second kind at this point.

The function has a point of removable discontinuity of the first kind at, and a point of discontinuity of the second kind at.

References:
O.I. Demons. Lectures on mathematical analysis. Part 1. Moscow, 2004.

A selection online calculators for a complete exploration of the function and plotting.

Find Function Scope

Calculate Parity of a Function

Calculation of the points of intersection of the graph with the axis (zeros of the function)

Find extrema of a function

Inflection points, convexity and concavity intervals

Plot function

This calculator is designed to find the breakpoints of a function online.

Discontinuity points of a function are points at which a function has a discontinuity, and the function at these points is not continuous.

There is a definite classification of points of discontinuity of a function. The discontinuity points of a function are divided into discontinuity points of the first kind and points of discontinuity of the second kind.

Discontinuity points of the first kind for x = a take place if there are left-sided and right-sided limits: lim (x → a-0) ⁡f (x) and lim (x → a + 0) ⁡f (x). These limits must be finite. If at least one of the one-sided limits is zero or infinity, then the function has discontinuity points of the second kind.

In order to find the breakpoints of a function online, you need to specify the function and the value of the argument.

To get the full course of the solution, click in the Step-by-step answer.

Explore a function, build a graph

Plan function research and plotting.

The answer means the following: even - the function is even, odd - the function is odd, neither even nor odd - the function is neither even nor odd.

3. Points of intersection of the graph of the function with the coordinate axes;

4. Continuity of function, break points;

5. Asymptotes of the graph of the function;

6. Intervals of monotony and critical points;

7. Bulge intervals and inflection points;

8. Making a schedule based on the research.

Online educational services: theory and practice

Solutions to typical problems - Mathematical analysis

Examine the function for continuity, determine the nature of the discontinuity.

Example 1 .

The function is not defined at the points, the first continuity condition has already been violated, therefore, at these points the function experiences a discontinuity.

To find out the nature of the discontinuity, it is necessary to calculate the one-sided limits in points.

Since the left limit at a point is equal to infinity, there is a discontinuity of the second kind in it.

Since the right limit at a point is equal to infinity, then there is a discontinuity of the second kind in it.

Example 2 The function is defined on the entire number line, but at the same time it is not continuous, since, i.e. the right and left limits at zero are not equal to each other and are not equal to the value of the function at zero, conditions 2 and 3 of continuity are violated. Since the right and left limits at zero exist and are finite, this is a discontinuity of the first kind.

Example 3 The function is undefined at zero, hence the break point.

Since and, this is a removable discontinuity, the function can be extended at zero “by continuity” by setting it equal to one.

Example 4

The function is elementary, therefore it is continuous in the domain of its definition. The domain of definition does not include points, therefore, they are the break points of the given function.

Let's define the type of break points.

Since, then the point is the point

discontinuity of the second kind of function.

The one-sided limits of a function at a point are equal, but the function at is not defined, therefore, it is a removable discontinuity point of the first kind.

Since the given function is even function, then it is obvious that

And it is the breaking point of the second kind of function.

To construct a sketch of the graph of a function, we investigate the behavior of the function when

and. Since the function is even, then

Let's build a sketch of the function graph.

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Some tasks can be solved online by entering numerical values, with detailed solution.

We construct (investigate) the graph of the function y = f (x), for this, define the function f (x)

Important: a should be less b, otherwise the graph will not be able to build. Watch the scale - if there is no graph in the picture, then it is worth varying the values a and b

Using the degree

(square and cube) and fractions

Using sine and cosine

Hyberbolic sine and cosine

Hyberbolic tangent and cotangent

Hyberbolic arcsine and arccosine

Hyberbolic arc tangent and arc cotangent

For periodic functions, the study of the function graph is carried out only on the period interval

Our calculator allows you to explore the graph of a function. But so far there is no way to find the scope of the function.

What this calculator can find:

Rules for entering expressions and functions

Expressions can consist of functions (designations are given in alphabetical order):

absolute (x) Absolute value x

(module x or | x |) arccos (x) Function - inverse cosine of x arccosh (x) Arccosine hyperbolic from x arcsin (x) Arcsine of x arcsinh (x) Arcsine hyperbolic of x arctg (x) Function - arctangent of x arctgh (x) Arctangent hyperbolic of x e e a number that is roughly 2.7 exp (x) Function - exponent from x(as e^x) log (x) or ln (x) Natural logarithm from x

(To obtain log7 (x), you must enter log (x) / log (7) (or, for example, for log10 (x)= log (x) / log (10)) pi The number is Pi, which is approximately 3.14 sin (x) Function - Sine of x cos (x) Function - Cosine of x sinh (x) Function - Sine hyperbolic from x cosh (x) Function - Cosine hyperbolic from x sqrt (x) Function - Square root from x sqr (x) or x ^ 2 Function - Square x tg (x) Function - Tangent of x tgh (x) Function - Tangent hyperbolic from x cbrt (x) Function - cube root of x floor (x) Function - rounding x downward (example floor (4.5) == 4.0) sign (x) Function - Sign x erf (x) Error function (Laplace or Probability Integral)

The following operations can be used in expressions:

Real numbers enter in the form 7.5 , not 7,5 2 * x- multiplication 3 / x- division x ^ 3- exponentiation x + 7- addition x - 6- subtraction

Test work RU - online calculators

Continuity and plotting of piecewise-defined functions is a complex topic. It is better to learn to build graphs directly in a practical lesson. This is mainly a continuity study.

It is known that elementary function(see p. 16) is continuous at all points at which it is defined. Therefore, the discontinuity of elementary functions possible only at points of two types:

a) at the points where the function is "redefined";

b) at points where the function does not exist.

Accordingly, only such points are checked in the study for continuity, which is shown in the examples.

For non-elementary functions, the study is more difficult. For example, a function (integer part of a number) is defined on the entire number axis, but breaks at every integer x... Such questions are beyond the scope of this manual.

Before studying the material, you should repeat in a lecture or a textbook what (what kind) are the break points.

Study of piecewise-defined functions for continuity

Function set piecewise, if it is given by different formulas in different parts of the domain of definition.

The main idea when examining such functions is to find out whether the function is specified at the points at which it is redefined, and how. Then it is checked whether the values ​​of the function to the left and to the right of such points are the same.

Example 1. Let us show that the function
continuous.

Function
is elementary and therefore continuous at the points at which it is defined. But, obviously, it is defined at all points. Consequently, it is continuous at all points, including for
as the condition requires.

The same is true for the function
, and at
it is continuous.

In such cases, continuity can only be broken where the function is overridden. In our example, this is the point
... Let's check it, for which we find the limits on the left and right:

The left and right limits are the same. It remains to find out:

a) is the function defined at the point itself
;

b) if so, does it match
with left and right limit values.

By condition, if
, then
... That's why
.

We see that (all are equal to the number 2). This means that at the point
function is continuous... So, the function is continuous on the whole axis, including the point
.

Solution Notes

a) It did not matter in the calculations, substitute we are in a specific formula number
or
... This is usually important when a division by an infinitesimal value is obtained, since it affects the sign of infinity. Here
and
responsible only for function selection;

b) as a rule, notation
and
are equal, the same applies to designations
and
(and is true for any point, not just for
). Further, for brevity, the designations of the form are used
;

c) when the limits on the left and right are equal, in order to check for continuity, in fact, it remains to see whether one of the inequalities not strict... In the example, this turned out to be the 2nd inequality.

Example 2. Let us investigate the continuity of the function
.

For the same reasons as in example 1, the continuity can be broken only at the point
... Let's check:

The limits on the left and right are equal, but at the very point
function is undefined (strict inequalities). It means that
- point removable gap.

"Removable gap" means that it is enough either to make any of the inequalities lax, or to come up with a single point
function, the value of which at
equals –5, or simply indicate that
so that the whole function
became continuous.

Answer: point
- point of removable gap.

Remark 1. In the literature, a removable gap is usually considered a special case of a type 1 gap, however, students are more often understood as a separate type of gap. In order to avoid misunderstandings, we will adhere to the 1st point of view, and the "irreparable" gap of the 1st kind will be specially stipulated.

Example 3. Let us check whether the function is continuous

At the point

The limits on the left and right are different:
... Whether or not the function is defined at
(yes) and if yes, then what is (equal to 2), point
–irreparable break point of the 1st kind.

At the point
going on final jump(from 1 to 2).

Answer: point

Remark 2. Instead of
and
usually write
and
respectively.

Available question: how the functions differ

and
,

as well as their schedules? Correct answer:

a) the 2nd function is not defined at the point
;

b) on the graph of the 1st function point
"Painted over", on the graph the 2nd is not ("punctured point").

Point
where the graph ends
, not shaded in both graphs.

It is more difficult to investigate functions that are differently defined on three plots.

Example 4. Is the function continuous
?

Just as in examples 1 - 3, each of the functions
,
and continuous on the entire numerical axis, including in the area where it is specified. Break is possible only at the point
or (and) at the point
where the function is overridden.

The problem breaks down into 2 subtasks: investigate for the continuity of a function

and
,

and point
not of interest to the function
and point
- for function
.

1st step. Checking the point
and function
(we don't write the index):

The limits are the same. By condition,
(if the limits on the left and right are equal, then in fact the function is continuous when one of the inequalities is not strict). So at the point
the function is continuous.

2nd step. Checking the point
and function
:

Insofar as
, point
Is the break point of the 1st kind, and the value
(and whether it exists at all) no longer plays a role.

Answer: the function is continuous at all points except the point
, where there is an unremovable break of the 1st kind - a jump from 6 to 4.

Example 5. Find the discontinuity points of a function
.

We act in the same way as in example 4.

1st step. Checking the point
:

a)
since to the left of
the function is constant and equal to 0;

b) (
Is an even function).

The limits are the same, but at
the function is not defined by condition, and it turns out that
- point of removable gap.

2nd step. Checking the point
:

a)
;

b)
- the value of the function does not depend on the variable.

The limits are different: , point
- point of irreparable rupture of the 1st kind.

Answer:
- point of removable gap,
Is a point of unremovable discontinuity of the 1st kind, at other points the function is continuous.

Example 6. Is the function continuous
?

Function
defined at
, therefore the condition
turns into a condition
.

On the other hand, the function
defined at
, i.e. at
... Hence, the condition
turns into a condition
.

It turns out that the condition must be met
, and the domain of the entire function is the segment
.

The functions themselves
and
are elementary and therefore continuous at all points at which they are defined - in particular, and for
.

It remains to check what is happening at the point
:

a)
;

Insofar as
, see if the function is defined at the point
... Yes, 1st inequality is relatively loose
, and that's enough.

Answer: the function is defined on the segment
and is continuous on it.

More complex cases, when one of the constituent functions is non-elementary or not defined at any point of its segment, are beyond the scope of the manual.

NF1. Plot function graphs. Pay attention to whether the function is defined at the point at which it is overridden, and if so, what is the meaning of the function (the word “ if"Omitted for brevity in the function definition):

1) a)
b)
v)
G)

2) a)
b)
v)
G)

3) a)
b)
v)
G)

4) a)
b)
v)
G)

Example 7. Let be
... Then on the site
build a horizontal line
, and on the site
build a horizontal line
... In this case, the point with coordinates
"Gouged out", and the point
"Painted over". At the point
a break of the first kind ("jump") is obtained, and
.

NF2. Investigate the continuity of functions, differently defined on 3 intervals. Plot graphs:

1) a)
b)
v)

G)
e)
e)

2) a)
b)
v)

G)
e)
e)

3) a)
b)
v)

G)
e)
e)

Example 8. Let be
... Location on
building a straight line
, for which we find
and
... Connect the dots
and
segment. We do not include the points themselves, since for
and
function is not defined by condition.

Location on
and
we circle the axis OX (on it
), but the points
and
"Gouged out". At the point
we obtain a removable discontinuity, and at the point
- a break of the 1st kind ("jump").

NF3. Plot function graphs and make sure they are continuous:

1) a)
b)
v)

G)
e)
e)

2) a)
b)
v)

G)
e)
e)

NF4. Make sure the functions are continuous and plot their graphs:

1) a)
b)
v)

2 a)
b)
v)

3) a)
b)
v)

NF5. Plot function graphs. Note the continuity:

1) a)
b)
v)

G)
e)
e)

2) a)
b)
v)

G)
e)
e)

3) a)
b)
v)

G)
e)
e)

4) a)
b)
v)

G)
e)
e)

5) a)
b)
v)

G)
e)
e)

NF6. Plot discontinuous functions. Notice the value of the function at the point where the function is being overridden (and whether it exists):

1) a)
b)
v)

G)
e)
e)

2) a)
b)
v)

G)
e)
e)

3) a)
b)
v)

G)
e)
e)

4) a)
b)
v)

G)
e)
e)

5) a)
b)
v)

G)
e)
e)

NF7. The same task as in NF6:

1) a)
b)
v)

G)
e)
e)

2) a)
b)
v)

G)
e)
e)

3) a)
b)
v)

G)
e)
e)

4) a)
b)
v)

G)
e)
e)

Continuity of a function at a point. Function y = f (x ) is called continuous

breakout at point x 0, if:

1) this function is defined in some neighborhood of the point x 0;

2) there is a limit lim f (x);

→ x 0

3) this limit is equal to the value of the function at the point x 0, i.e. limf (x) = f (x 0).
		x → x0
The last condition is equivalent to lim		y = 0, where x = x - x 0 - at-
	x → 0
argument expansion, y = f (x 0 +	x) - f (x 0) - function increment corresponding to
incrementing argument	x, i.e. function	f (x) is continuous at the point x 0

if and only if at this point the infinitesimal increment of the argument corresponds to the infinitesimal increment of the function.

One-way continuity. The function y = f (x) is called continuous

left at the point x 0 if it is defined on some half-interval (a; x 0]

and lim f (x) = f (x 0).

x → x0 - 0

A function y = f (x) is called continuous on the right at a point x 0 if it is

is defined on some half-interval [x 0; a) and limf (x) = f (x 0).

x → x0 + 0

Function y = f (x)		is continuous at the point x 0			if and only if she
continuous
lim f (x) = limf (x) = limf (x) = f (x 0).
x → x0 + 0	x → x0 - 0	x → x0

Continuity of a function on a set. The function y = f (x) is called

continuous on the set X if it is continuous at every point x of this set. Moreover, if the function is defined at the end of a certain interval of the numerical axis, then by continuity at this point we mean continuity on the right or on the left. In particular, the function y = f (x) is called

discontinuous on the segment [a; b] if it

1) continuous at every point of the interval(a; b);

2) is continuous on the right at the point a;

3) is continuous on the left at the point b.

Function breakpoints. The point x 0, belonging to the domain of definition of the function y = f (x), or being the boundary point of this domain, is called

the breakpoint of this function if f (x) is not continuous at this point.

Breakpoints are subdivided into breakpoints of the first and second kind:

1) If there are finite limits lim f (x) = f (x 0 - 0) and

x → x0 - 0

	f (x) = f (x 0 + 0), and not all three numbers f (x 0 - 0), f (x 0 + 0),		f (x 0) are equal
x → x0 + 0
each other, then x 0		is called a discontinuity point of the first kind.
	In particular, if the left and right limits of the function at the point x 0		equal between
	myself, but	are not equal to the value of the function at this point:

f (x0 - 0) = f (x0 + 0) = A ≠ f (x0), then x 0 is called a removable discontinuity point.

In this case, putting f (x 0) = A, we can modify the function at the point x 0

so that it becomes continuous ( extend the function by continuity). The difference f (x 0 + 0) - f (x 0 - 0) is called jump function at point x 0.

The jump of the function at the point of a removable discontinuity is zero.

2) Discontinuity points that are not discontinuity points of the first kind are called break points of the second kind... At discontinuity points of the second kind, at least one of the one-sided limits f (x 0 - 0) and f (x 0 + 0) does not exist or is infinite.

Properties of functions continuous at a point.

		f (x)	and g (x) are continuous at the point x 0, then the functions
f (x) ± g (x),	f (x) g (x) and	f (x)	(where g (x) ≠ 0) are also continuous at the point x.
		g (x)

2) If the function u (x) is continuous at the point x 0, and the function f (u) is continuous

at the point u 0 = u (x 0), then the complex function f (u (x)) is continuous at the point x 0.

3) All basic elementary functions (c, x a, a x, loga x, sinx, cosx, tgx, ctgx, secx, cosecx, arcsinx, arccosx, arctgx, arcctgx) are continuous in each

to the point of their domains of definition.

From properties 1) –3) it follows that all elementary functions (functions obtained from basic elementary functions using a finite number of arithmetic operations and composition operations) are also continuous at each point of their domains of definition.

Properties of functions that are continuous on a segment.

1) (theorem about intermediate values) Let the function f (x) be defined

on and is continuous on the segment [a; b]. Then for any number C enclosed

between the numbers f (a) and f (b), (f (a)< C < f (b )) найдется хотя бы одна точкаx 0 [ a ;b ] , такая, чтоf (x 0 )= C .

2) (Bolzano - Cauchy theorem

is continuous on the segment [a; b] and takes values ​​of different signs at its ends.

Then there is at least one point x 0 [a; b] such that f (x 0) = 0.

3) (1st Weierstrass theorem) Let the function f (x) be defined and

is continuous on the segment [a; b]. Then this function is bounded on this segment.

4) (2nd Weierstrass theorem) Let the function f (x) be defined and

jerky on the segment		[a; b]. Then this function reaches on the segment [a; b]
the greatest		the smallest	values, i.e.	exists
x1, x2 [a; b],		for any	points x [a; b]	fair	inequalities

f (x 1) ≤ f (x) ≤ f (x 2).

Example 5.17. Using the definition of continuity, prove that the function y = 3x 2 + 2x - 5 is continuous at an arbitrary point x 0 of the numerical axis.

Solution: Method 1: Let x 0 be an arbitrary point of the numerical axis. You-

we first calculate the limit of the function f (x) as x → x 0, applying theorems on the limit of the sum and product of functions:

lim f (x) = lim (3x 2 + 2x - 5) = 3 (limx) 2 + 2 limx - 5 = 3x 2							− 5.
x → x0	x → x0	x → x0	x → x0
Then we calculate the value of the function at the point x: f (x) = 3x 2							− 5 .
Comparing the results obtained, we see				lim f (x) = f (x 0), which, according to
				x → x0

definition and means the continuity of the considered function at the point x 0.

Method 2: Let		x - increment of the argument at point x 0. We find the corresponding
corresponding	increment			y = f (x0 + x) - f (x0) =
3 (x + x) 2 + 2 (x + x) - 5− (3x 2 + 2x - 5)
6 x x + (x) 2	2x = (6x + 2) x + (x) 2.
Let us now calculate the limit of the increment of the function when the increment of the argument is
		seeks

	y = lim (6x + 2)		x + (x) 2 = (6x + 2) lim		x + (limx) 2 = 0.
x → 0	x → 0			x → 0	x → 0

Thus, lim y = 0, which means, by definition, the continuity

x → 0

functions for any x 0 R.

Example 5.18. Find the discontinuity points of the function f (x) and determine their genus. V

in the case of a removable discontinuity, define the function by continuity:

1) f (x) = 1 - x 2 for x< 3;

5x for x ≥ 3

2) f (x) = x 2 + 4 x + 3;

x + 1

	f (x) =
		x4 (x− 2)
	f (x) = arctg
			(x - 5)

Solution: 1) The scope of this function is the whole number -

axis (−∞; + ∞). The function is continuous on the intervals (−∞; 3), (3; + ∞). The gap is possible only at the point x = 3, at which the analytical setting of the function changes.

Find the one-sided limits of the function at the specified point:

f (3− 0) = lim (1− x 2) = 1− 9 = 8;

x → 3 −0

f (3+ 0) = lim 5x = 15.

x → 3 +0

We can see that the left and right limits are finite, so x = 3
break I			f (x). Function jump in
f (3+ 0) - f (3− 0) = 15− 8 = 7.
		f (3) = 5 3 = 15 = f (3+ 0), therefore at the point		x = 3

f (x) is right continuous.

2) The function is continuous on the entire number axis, except for the point x = - 1, in which it is not defined. We transform the expression for f (x) by expanding the numerator

fractions for factors:		f (x) =			4 x +3			(x + 1) (x + 3)	X + 3 for x ≠ - 1.
					x + 1			x + 1
Find the one-sided limits of the function at the point x = - 1:
	f (x) = lim	f (x) = lim (x + 3) = 2.
x → −1 −0	x → −1 +0		x → −1

We found out that the left and right limits of the function at the point under study exist, are finite and are equal to each other, therefore x = - 1 is the point of removable

the straight line y = x + 3 with the "punctured" point M (- 1; 2). To make the function indispensable

discontinuous, one should put f (- 1) = f (- 1 - 0) = f (- 1+ 0) = 2.

Thus, by extending the definition of f (x) by continuity at the point x = - 1, we have obtained a function f * (x) = x + 3 with the domain of definition (−∞; + ∞).

3) This function is defined and continuous for all x except points

x = 0, x = 2, where the denominator of the fraction vanishes.

Consider a point x = 0:

Since in a sufficiently small neighborhood of zero the function takes only

	negative values, then f (- 0) = lim			= −∞ = f (+0)	Those. point
			(x - 2)
	x → −0
	x = 0 is a discontinuity point of the second kind of the function		f (x).

Consider now the point x = 2:

The function takes negative values ​​near to the left of the considered

	given point and positive - to the right, therefore				f (2− 0) =			= −∞,
							x4 (x− 2)
						x → 2 −0
	f (2+ 0) = lim			= + ∞. As in the previous case, at the point x = 2
			(x - 2)
	x → 2 +0

tion has neither left nor right finite limits, i.e. suffers a break of the second kind at this point.

x = 5.

f (5− 0) = lim arctan

π, f (5+ 0) = lim arctan

x = 5

(x - 5)

(x - 5)

x → 5 −0

x → 5 +0

ka break

f (5+ 0) - f (5− 0) =

π − (−

π) = π (see Figure 5.2).

Tasks for independent solution

5.174. Using only the definition, prove the continuity of the function f (x) in

each point x 0 R:

a) f (x) = c = const;		b) f (x) = x;
c) f (x) = x 3;		d) f (x) = 5x 2 - 4x + 1;
e) f (x) = sinx.
5.175. Prove that the function	f (x) = x 2	1 for x ≥ 0,	is continuous on
		1 at x< 0
the entire number axis. Plot this function.
5.176. Prove that the function	f (x) = x 2	1 for x ≥ 0,	is not continuous
		0 for x< 0

at the point x = 0, but is continuous on the right at this point. Plot the function f (x).

breakout point x =

But it is continuous on the left at this point. Build a graph

function f (x).

5.178. Plot function graphs

a) y =

x + 1

b) y = x +

x + 1

x + 1

x + 1

Which of the conditions of continuity at the points of discontinuity of these functions are satisfied, and which are not?

5.179. Specify function break point

sin x		For x ≠ 0
		at x = 0

Which of the conditions of continuity at this point are satisfied, and which are not?