Theorem

Let the function be defined, strictly monotonic and continuous on a certain interval, and let be the set of its values. Then the inverse function on the set unambiguous, strictly monotonic and continuous.

Proof

For definiteness, let the function increases by , i.e. for any , satisfying the condition , the inequality holds:

(), ().

1. Let us prove the uniqueness of the inverse function.

Uniqueness of the inverse function follows from the fact that due to the increase in the function the inequality is valid:

At ,

and, therefore, to everyone matches a single value .

2. Let us now prove that the inverse function increases by .

Indeed, if , then (And ), because if it were , then from increasing it should be that , which would contradict the assumption . Thus, the fact of strict monotonicity of the inverse function installed.

3. And finally, we prove that the inverse function is continuous on .

Because monotonically increases on the set, then it is bounded and takes the largest and smallest values ​​on the set. The set is an interval with ends and , where , .

Let , . Let us first consider the case when . In this case, the point is obviously an interior point of the interval.

Let's choose a value such that And , and put And . Then, due to increasing we get:

.

Let's take now such that the following inequalities are satisfied:

And .

Then, if satisfies the inequalities

,

That ,

and, therefore, due to the increase we have:

Considering that and

we get: provided
.

Thus, it has been proven that for any sufficiently small exists such that for all satisfying the inequality , the inequality holds , i.e. inverse function is continuous at the point. But - arbitrary point of the interval . So the inverse function continuous on .

If or , then using similar reasoning we can prove the continuity on the right at point and on the left at point . So, the fact of continuity of the inverse function not proven.

In case of decreasing function the proof of the theorem is carried out similarly.

Module

Topic No. 5

Continuity of main

Elementary functions. Uniform continuity of a function on a set

Lecture No. 17

1. Continuity of functions: ; ; ; ;
;
;
; ; ; .

2. Exponential function in the set of rational numbers.

3. Exponential function in the set of real numbers.

ﻫ Continuity of elementary functions

1. Prove that the function ,

Proof

1) Choose an arbitrary point R, because determined on R.

2) For this point R Let's determine the limit and value of the function at the point:

A)
,

b) .

3) Therefore, , i.e. function continuous at any point .

,

is continuous at every point on the number line.

5) The proof is carried out on the basis of definition No. 1 of the continuity of a function at a point. Etc.

2. Prove that the function is continuous at any point on the number line except zero, i.e. R\0.

Proof

R\0, since the function is defined on R\0, and define the function increment in it:

.

3) Calculate the limit

,

because . So the function

4) Since the point was chosen arbitrarily, then the function continuous at any point R\0. Etc.

3. Prove that the function is continuous at any point in the set of real numbers.

Proof

1) The domain of definition of a function is the set of real numbers.

2) Choose an arbitrary point R and define the increment of the function in it:

3) Calculate the limit
. So the function continuous at any point .

4) Since the point was chosen arbitrarily, then the function continuous at any point in the set of real numbers .

5) The proof was carried out on the basis of definition No. 5 of the continuity of a function at a point in the language of increments. Etc.

4. Prove that the function is continuous at any point in the set R.

Proof

The proof follows from the theorem on the continuity of an algebraic sum, the product and quotient of continuous functions and the continuity of a function at any point on the number axis. Etc.

5. Prove that the function
is continuous at any point in the set of real numbers, except for those points where the denominator of the fraction vanishes.

Proof

The proof follows from the theorem on the continuity of an algebraic sum, product and quotient of continuous functions and continuity of functions

And .

This means that the given function is continuous at any point in the set R, excluding those points at which the denominator is zero. Etc.

6. Prove that the function is continuous at any point on the number line.

Proof

1) We will carry out the proof on the basis of definition No. 5 of the continuity of a function at a point in the language of increments.

2) Function defined at any point on the number line.

3) Choose an arbitrary point R and determine the increment of the function at this point:

4) Let's calculate the limit on the increment of the function:

So the function is continuous at an arbitrary point.

5) Since the point was chosen arbitrarily, then the function is continuous at any point on the number line. Etc.

7. The continuity of the function is proved in a similar way at any point on the number axis. Conduct the proof yourself.

8. From the continuity of functions And at any point on the number line, according to the theorem on the continuity of the quotient of continuous functions at a point, the continuity of the function follows

A) ; at all points on the number line, except points

, - any integer;

b) as well as continuity of the function And at all points except points , where is any integer.

9. Prove that the function continuous on the entire number line.

Proof

1) On the interval function looks like , because . And this function is continuous at every point on the number line.

2) On the interval function looks like , because . And this function is continuous as the product of two continuous functions and .

3) It remains to establish the continuity of the function at the point .

4) To do this, we calculate one-sided limits at the point :

A) ; b) .

5) Since
And , then the function continuous at a point . And, therefore, it is continuous along the entire number line.

Conclusion:

1. All the functions considered are continuous in their domains of existence.

2. Based on the theorems of continuity of the sum, difference, product and quotient of continuous functions, it can be argued that functions obtained using a finite number of arithmetic operations on continuous functions are also continuous functions in the domain of their existence.

Exponential function in multirational numbers

Definition 1. Let , then for any rational number the value will be determined. This defines the function. This function is called exponential on the set of rational numbers.

Properties of the Exponential Function

I.... i.e. , where , .1.Let . Then:a) if , then ;b) if , then .2.a) ;b) ;c) .3. .4. .5. , for any rational number: .

Document: 5th property1. If and , then by virtue of the first property: .

2. Since , a , then .3. Based on the second property: , and , therefore, .4. The inequality for is proved in a similar way.

II. Lemma 1. Let . Then it exists that for all rational numbers satisfying the inequality the following inequality holds: . Help: at .

Document: I.1. Let .

2. Since , then: and .3. Since , then based on the first property, therefore, the two double inequalities can be rewritten as follows:

4. Let be a rational number such that , that is, .5. Then, based on the first property of the exponential function, we can write: or or .II. The lemma is obvious. III. When the lemma is proved in a similar way, only in accordance with the inequality of the first property the sign must be replaced with the opposite one (case 1b).

2 Exponential function in the set of real numbers

Definition 2. Let a be an arbitrary real number, that is, . Let be a sequence of rational numbers converging to . Obviously, such a sequence always exists. Then , , always exists and does not depend on the choice of sequence.

The case is not of interest for study, since .

Theorem 1. An exponential function in the set of real numbers , , has the following properties: 1) continuous at every point on the number line; 2) for strictly increases, and for strictly decreases on the entire number line; 3) , ; 4) , ; 5) a) at ;b) at ;6)a) at ;b) at .

Document: 1st property1. It is known that: .2. This statement is also true for real numbers.3. Let be an arbitrary real number, , and , , be an exponential function in the set of real numbers.4. Let's find the increment of the function at a point when the argument changes to: .

5. According to the lemma for the exponential function in the set of rational numbers: , satisfying the inequality ), the following inequality is satisfied: , and for , .6. Let's multiply both sides of the inequality in point 5 by a positive number: .7. Let's compare the increment of the function and the last inequality, it is obvious that for , i.e. , that is, based on definition No. 5 of the continuity of a function at a point, the function is continuous at point .8. Since the point was chosen arbitrarily, the function is continuous at any point on the number line.

Document: 2nd property1. Let for definiteness and .2. Due to the density of rational numbers in the set of real numbers, there exist rational numbers and such that

3. Let us choose some two sequences of rational numbers and so that and , and so that for .4. When, based on the first property of the exponential function in the set of rational numbers, we can write:

5. Let's move on to the limit at (in exponents) in the last inequality: ; ; ; , therefore, based on the definition of the exponential function in the set of real numbers: ; ; .6. The inequality in point 4 will take the form: or at .7. And based on the definition of an increasing function at , therefore, strictly increases at .

Note 1. The case is treated similarly.

Note 2. The function graphs look like:

Document: 3rd properties1. Let there be sequences of rational numbers such that , and, therefore, based on the theorem on the limit of the sum of two convergent sequences.2. Then, by virtue of the definition of the exponential function in the set of real numbers, .3. By property No. 2 of the exponential function in the set of rational numbers, therefore, a>0.

Corollary 1. For any real numbers the following equality holds: , therefore, . 2. Therefore.

Document: 4th property

I.1. Let be a positive integer, i.e. .2. Let us once again apply property No. 2 of the exponential function in the set of rational numbers: Therefore, .

II.1. Let , , where is a positive integer, .2. Let us prove that: , i.e. which is the root of the power of the number: .3. Based on equality and by definition of a root: if , then . Or . Hence, .

III.1. Let , , where .2. Based on what was previously proven, we can write: . Hence, .

IV. Let now, then. Hence, .

V. It is obvious that .Conclusion: thus, it is proven that , : .

VI.1. Let .2. Consider an arbitrary sequence of rational numbers converging to: .3. Then, by virtue of equality, the following will occur:

4. Since , then, in accordance with the definition of an exponential function in the set of real numbers, we can write: a) ; b) , since .5. Let's move on to the limit in the equality of point 3 at:

6. In accordance with paragraph 4, we rewrite the written equality: , .

Document: 5th propertyI.1. Let us prove that for .2. Let .3. Then , .4.Since , then (in accordance with Bernoulli’s inequality).5. For , that .6. If , a , then at , therefore, the definition of an infinitely large function at (the infinite limit of a function at infinity) is equivalent to .

II.1. If , then .2 is satisfied. Since, then, i.e. .3. Therefore .4. If and , then , therefore, especially on the basis of the compressed variable theorem.

Document: 6th property1. At .2. When .The proof is carried out similarly to the proof of property No. 5.

Module

CONTINUITY OF THE INVERSE FUNCTION

Definition 1. Let f– correspondence between sets X And Y. The set of all pairs (( y,x)| (x,y)Î f) is called the inverse correspondence for compliance f and is designated f –1 .

Definition 2. If matches f And f–1 are functions, then the function f called reversible, A f –1 –reverse for function f .

Functions f And f–1 are mutually inverse, because ( f –1) –1 = f, and the display

f: X Y is one-to-one.

Properties of mutually inverse functions:

1. D(f -1) = E(f), E(f -1) = D(f).

2. f –1 (f(x)) = x "xОD( f); f(f –1 (y)) = y "yÎ E(f).

3. Function graphs f And f–1 – symmetrical about a straight line y = x.

Let us accept the following theorem without proof

Theorem 1. If the function f is a one-to-one mapping of the domain of definition D(f) to the range of values E(f), then its inverse correspondence f–1 – function.

Theorem 2 ( on the existence and continuity of the inverse function). Let the function f be strictly increasing (decreasing) and continuous on the domain of definition D(f), which is an interval. Then the inverse correspondence f –1 is a function of increasing (decreasing) and continuous in its domain of definition D(f –1 ) = E(f), which is also the interval.

Note, that according to the corollary of the Bolzano-Cauchy theorem II, the range of values ​​of a function continuous on an interval E(f) = D(f–1) – interval.

Proof We will carry out this for an increasing function in 3 stages.

Stage 1. Let f– increasing, let us prove that f –1 – function, i.e. we will show that to everyone

yÎ D(f –1) = E(f) corresponds to a single value XÎ E(f–1) = D( f).

Let us assume the opposite, that for some y oÎ E(f) correspond to two x 1, x 2Î D(f) such that f(x 1) = y o і f(x 2)= y o, But x 1 ≠ x 2. Let for definiteness x 1< x 2. From the condition of increasing function f follows that f(x 1) < f(x 2)Û y o< y o , but this is impossible.

Stage 2.Let us prove that f –1 – increasing function in domain D(f –1) = E(f). In abundance E(f) let's take any at 1 And at 2 such that at 1 < at 2 and show that f –1 (at 1)< f –1 (at 2).

Let's assume the opposite: f –1 (at 1) ³ f –1 (at 2). Due to the increasing function f we will mark

f(f –1 (y 1)) ³ f(f –1 (at 2)) Þ y 1 ³ y 2, which contradicts the condition at 1 < at 2. This proves the increase in the function f –1 .

Stage 3.Let us say that the function f –1 continuous on E(f).

We have proven that f–1 – increasing over the interval E(f) function, set of its values E(f -1) = D(f) according to the conditions of the theorem – an interval. Then by T.2 §4 f–1 – continuous function on E(f). ◄

Example 1. Find the inverse function of a function function f (X) = 2x - 4.

Solution. Function f (X) = 2x- 4 – continuous and increasing by D(f) = R. According to T. 2, there is an inverse function, which is also continuous and increasing by E(f) = R. Let's find the formula for the function f –1 (at), for this we express X = at/2 + 2, or

y = x/2 + 2 (X And at swapped places).

Example 2. Find the inverse function of a function

and build its graph.

Solution. D(f) = R – gap. Let us rewrite function (1) in the form Þ Þ e y-e–y= 2xÞ e y - 1/e y= 2x Þ e 2y - 2xe y- 1 = 0 ½ denote e y = t> 0½Þ

Theorem(on the existence and continuity of the inverse function). Let a continuous increasing (decreasing) function y=f(x) be defined on the interval (a,b). Let's denote

Then on the interval (A, B) an inverse function is defined, which increases (decreases) on this interval and is continuous at each point of this interval.

QUESTION No. 22: Differentiable functions. Differentiability criterion

Definition . Function f, defined in the neighborhood of a point x, called differentiable at this point, if the formula is true

f (x stroke+ ▲ x stroke)- f (x prime)=S Ai▲ xi+Sai (▲ x stroke)▲ xi (3)

Where i A are numbers, and functions ai (▲ x stroke) satisfy the condition

ai (▲ x stroke)→ 0 (i=1,2 ,…, n) at ▲ x→0 . (4)

Theorem . Let the function f differentiable at the point x . Then at this point it has partial derivatives and the equalities are satisfied

(df(x prime))/ (dxi)= Ai(i=1,2,…,n).

Proof . From formula (3) it follows that

(f (x1 ,...,xi-1 ,xi+▲ xi ,xi+1,...,xn)- f (x1 ,...,xi-1 ,xi ,xi+1 ,..., x n))/(▲Xi)= Ai+αi(▲Xi).

Passing to the limit at ▲x i → 0, we obtain equality (5).

Theorem (sufficient conditions for differentiability of a function). If the function f has partial derivatives with respect to all variables in some neighborhood of the point x stroke , and all these partial derivatives are continuous in the very

point x stroke , then the specified function is differentiable at this point.

Theorem ( criterion for differentiability of a function). Function f(x), defined in the neighborhood of the point x, is differentiable at this point if and only if the derivative exists f׳( x). Wherein F= f׳( x).

Proof . Let there be a derivative f׳( x). Let's denote

a(t) =(((f(t)-f(x)) /(t-x)) - f׳(x)

f(t) =f (x)+ (t-x) f׳(x)+ (t-x)α (t),(α(t)→0). (2)

Let now equality (1) be satisfied. Then

((f(t)-f(x)) /(t-x) = F+ α(t) ,limα(t) = 0.

Therefore, there is a derivative f׳( x)= F.

QUESTION No. 23: Derivative of the sum, difference, product and quotient of two functions.

Derivative of sum and difference

Let functions f(x) and g(x) be given whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:

1. (f + g)’ = f ’ + g ’

2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, (f + g + h)’ = f’ + g’ + h’.

Strictly speaking, there is no concept of “subtraction” in algebra. There is a concept of “negative element”. Therefore, the difference f − g can be rewritten as the sum f + (−1) g, and then only one formula remains - the derivative of the sum.

· Task . Find the derivatives of the functions: f(x) = x 2 + sin x; g(x) = x 4 + 2x 2 − 3.

Solution. The function f(x) is the sum of two elementary functions, therefore:

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x + cos x;

We reason similarly for the function g(x). Only there are already three terms (from the point of view of algebra):

g '(x) = (x 4 + 2x 2 − 3)' = (x 4 + 2x 2 + (−3))' = (x 4)' + (2x 2)' + (−3)' = 4x 3 + 4x + 0 = 4x · (x 2 + 1).

Answer:
f ’(x) = 2x + cos x;
g ’(x) = 4x · (x 2 + 1).

Derivative of the product

The derivative of a product is calculated using a completely different formula. Namely:

(f g) ’ = f ’ g + f g ’

The formula is simple, but it is often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

· Task . Find the derivatives of the functions: f(x) = x 3 · cos x; g(x) = (x 2 + 7x − 7) e x .

Solution

Answer:
g ’(x) = x(x + 9) e x .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

Derivative of the quotient

If there are two functions f(x) and g(x), and g(x) ≠ 0 on the set we are interested in, we can define a new function h(x) = f(x)/g(x). For such a function you can also find the derivative:

Not weak, huh? Where did the minus come from? Why g 2? And like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.

Task. Find the derivatives of the functions: f(x) = x 3 · cos x; g(x) = (x 2 + 7x − 7) e x .

Solution. The function f(x) is the product of two elementary functions, so everything is simple:

f '(x) = (x 3 cos x)' = (x 3)' cos x + x 3 (cos x)' = 3x 2 cos x + x 3 (− sin x) = x 2 · (3cos x − x · sin x)

The function g(x) has a slightly more complicated first factor, but this does not change the general scheme. Obviously, the first factor of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:

g '(x) = ((x 2 + 7x − 7) e x)' = (x 2 + 7x − 7)' e x + (x 2 + 7x − 7) (e x)' = (2x + 7 ) · e x + (x 2 + 7x − 7) · e x = e x · (2x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x + 9) · e x .

Answer:
f ’(x) = x 2 · (3cos x − x · sin x);
g ’(x) = x(x + 9) e x