Definition of the inverse function and its properties: lemma on the mutual monotonicity of the direct and inverse functions; symmetry of graphs of direct and inverse functions; theorems on the existence and continuity of the inverse function for a function that is strictly monotonic on a segment, interval and half-interval. Examples of inverse functions. An example of solving a problem. Proofs of properties and theorems.

Content

See also: Definition of a function, upper and lower bounds, monotonic function.

Definition and properties

Definition of an inverse function
Let a function have a domain of definition X and a set of values ​​Y. And let it have the property:
for everyone.
Then for any element from the set Y one can associate only one element of the set X for which . This correspondence defines a function called inverse function To . The inverse function is denoted as follows:
.

From the definition it follows that
;
for everyone;
for everyone.

Property of symmetry of graphs of direct and inverse functions
The graphs of direct and inverse functions are symmetrical with respect to the straight line.

Theorem on the existence and continuity of an inverse function on an interval
Let the function be continuous and strictly increasing (decreasing) on ​​the segment. Then the inverse function is defined and continuous on the segment, which strictly increases (decreases).

For an increasing function. For decreasing - .

Theorem on the existence and continuity of an inverse function on an interval
Let the function be continuous and strictly increasing (decreasing) on ​​an open finite or infinite interval. Then the inverse function is defined and continuous on the interval, which strictly increases (decreases).

For an increasing function.
For decreasing: .

In a similar way, we can formulate the theorem on the existence and continuity of the inverse function on a half-interval.

If the function is continuous and strictly increases (decreases) on the half-interval or , then on the half-interval or the inverse function is defined, which strictly increases (decreases). Here .

If strictly increasing, then the intervals and correspond to the intervals and . If strictly decreasing, then the intervals and correspond to the intervals and .
This theorem is proven in the same way as the theorem on the existence and continuity of an inverse function on an interval.

Examples of inverse functions

arcsine

Graphs y = sin x and inverse function y = arcsin x.

Consider the trigonometric function sinus: . It is defined and continuous for all values ​​of the argument, but is not monotonic. However, if you narrow the scope of definition, you can identify monotonous areas. So, on the segment , the function is defined, continuous, strictly increasing and takes values ​​from -1 to +1 . Therefore, it has an inverse function on it, which is called arcsine. The arcsine has a domain of definition and a set of values.

Logarithm

Graphs y = 2 x and inverse function y = log 2 x.

The exponential function is defined, continuous and strictly increasing for all values ​​of the argument. Its value set is an open interval. The inverse function is the logarithm to base two. It has a domain of definition and a set of meanings.

Square root

Graphs y = x 2 and inverse function.

The power function is defined and continuous for all. The set of its values ​​is the half-interval. But it is not monotonic for all values ​​of the argument. However, on the half-interval it is continuous and increases strictly monotonically. Therefore, if we take the set as the domain of definition, then there is an inverse function called the square root. An inverse function has a domain and a set of values.

Example. Proof of the existence and uniqueness of a root of degree n

Prove that the equation , where n is a natural number, is a real non-negative number, has a unique solution on the set of real numbers, . This solution is called the nth root of a. That is, you need to show that any non-negative number has a unique root of degree n.

Consider a function of the variable x:
(P1) .

Let us prove that it is continuous.
Using the definition of continuity, we show that
.
We apply Newton's binomial formula:
(P2)
.
Let's apply the arithmetic properties of function limits. Since , then only the first term is nonzero:
.
Continuity has been proven.

Let us prove that function (A1) strictly increases as .
Let's take arbitrary numbers connected by inequalities:
, , .
We need to show that . Let's introduce variables. Then . Since , then from (A2) it is clear that . Or
.
Strict increase has been proven.

Let's find the set of values ​​of the function at .
At point , .
Let's find the limit.
To do this, we apply Bernoulli's inequality. When we have:
.
Since , then and .
Applying the property of inequalities for infinitely large functions, we find that .
Thus, , .

According to the inverse function theorem, the inverse function is defined and continuous on an interval. That is, for anyone there is a unique one that satisfies the equation. Since we have , this means that for any , the equation has a unique solution, which is called a root of degree n of the number x:
.

Proofs of properties and theorems

Proof of the lemma on the mutual monotonicity of direct and inverse functions

Let a function have a domain of definition X and a set of values ​​Y. Let us prove that it has an inverse function. Based on , we need to prove that
for everyone.

Let's assume the opposite. Let there be numbers, so that . Let it be so. Otherwise, let's change the notation so that it is . Then, due to the strict monotonicity of f, one of the inequalities must be satisfied:
if f is strictly increasing;
if f is strictly decreasing.
That is . A contradiction arose. Therefore, it has an inverse function.

Let the function be strictly increasing. Let us prove that the inverse function is also strictly increasing. Let us introduce the following notation:
. That is, we need to prove that if , then .

Let's assume the opposite. Let it be, but.

If, then. This case disappears.

Let . Then, due to the strict increase of the function , , or . A contradiction arose. Therefore, only chance is possible.

The lemma is proven for a strictly increasing function. This lemma can be proven in a similar way for a strictly decreasing function.

Proof of the property about the symmetry of the graphs of direct and inverse functions

Let be an arbitrary point on the graph of a direct function:
(2.1) .
Let us show that a point symmetrical to point A relative to a straight line belongs to the graph of the inverse function:
.
From the definition of the inverse function it follows that
(2.2) .
Thus, we need to show (2.2).

Graph of the inverse function y = f -1(x) is symmetrical to the graph of the direct function y = f (x) relative to the straight line y = x.

From points A and S we lower perpendiculars to the coordinate axis. Then
, .

Through point A we draw a line perpendicular to line . Let the lines intersect at point C. We construct a point S on a straight line so that . Then point S will be symmetrical to point A relative to the straight line.

Consider triangles and . They have two sides of equal length: and , and equal angles between them: . Therefore they are congruent. Then
.

Consider a triangle. Since then
.
The same applies to the triangle:
.
Then
.

Now we find and:
;
.

So, equation (2.2):
(2.2)
is satisfied, since , and (2.1) is satisfied:
(2.1) .

Since we chose point A arbitrarily, this applies to all points on the graph:
all points on the graph of a function, symmetrically reflected with respect to the straight line, belong to the graph of the inverse function.
Next we can change places. As a result we get that
all points of the graph of a function, symmetrically reflected with respect to a straight line, belong to the graph of the function.
It follows that the graphs of functions and are symmetrical with respect to the straight line.

The property has been proven.

Proof of the theorem on the existence and continuity of the inverse function on an interval

Let denote the domain of definition of the function - the segment.

1. Let us show that the set of function values ​​is the segment:
,
Where .

Indeed, since the function is continuous on the segment, then, according to Weierstrass’s theorem, it reaches a minimum and a maximum on it. Then, by the Bolzano-Cauchy theorem, the function takes all values ​​from the segment. That is, for anyone exists, for which. Since there is a minimum and a maximum, the function takes on the segment only values ​​from the set.

2. Since the function is strictly monotonic, then according to the above, there is an inverse function, which is also strictly monotonic (increases if it increases; and decreases if decreases). The domain of the inverse function is the set, and the set of values ​​is the set.

3. Now we prove that the inverse function is continuous.

3.1. Let there be an arbitrary internal point of the segment: . Let us prove that the inverse function is continuous at this point.

Let the point correspond to it. Since the inverse function is strictly monotonic, that is, the interior point of the segment:
.
According to the definition of continuity, we need to prove that for any there is a function such that
(3.1) for everyone.

Note that we can take it as small as we like. Indeed, if we have found a function for which inequalities (3.1) are satisfied for sufficiently small values ​​of , then they will automatically be satisfied for any large values ​​of , if we put at .

Let us take it so small that the points and belong to the segment:
.
Let us introduce and arrange the notation:



.

Let us transform the first inequality (3.1):
(3.1) for everyone.
;
;
;
(3.2) .
Since it is strictly monotonic, it follows that
(3.3.1) , if it increases;
(3.3.2) , if it decreases.
Since the inverse function is also strictly monotonic, inequalities (3.3) imply inequalities (3.2).

For any ε > 0 there is δ, so |f -1 (y) - f -1 (y 0) |< ε for all |y - y 0 | < δ .

Inequalities (3.3) define an open interval, the ends of which are distant from the point at distances and . Let there be the smallest of these distances:
.
Due to the strict monotonicity of , , . Therefore and. Then the interval will lie in the interval defined by inequalities (3.3). And for all values ​​belonging to it, inequalities (3.2) will be satisfied.

So we found that for small enough , there is , so that
at .
Now let's change the notation.
For small enough, there is such a thing, so
at .
This means that the inverse function is continuous at interior points.

3.2. Now consider the ends of the domain of definition. Here all the reasoning remains the same. You just need to consider one-sided neighborhoods of these points. Instead of a dot there will be or, and instead of a dot - or.

So, for an increasing function , .
at .
The inverse function is continuous at the point, since for any sufficiently small there is , so that
at .

For a decreasing function, .
The inverse function is continuous at the point, since for any sufficiently small there is , so that
at .
The inverse function is continuous at the point, since for any sufficiently small there is , so that
at .

The theorem has been proven.

Proof of the theorem on the existence and continuity of the inverse function on an interval

Let denote the domain of definition of the function - an open interval. Let be the set of its values. According to the above, there is an inverse function that has a domain of definition, a set of values ​​and is strictly monotonic (increases if it increases and decreases if it decreases). It remains for us to prove that
1) the set is an open interval, and that
2) the inverse function is continuous on it.
Here .

1. Let us show that the set of function values ​​is an open interval:
.

Like any non-empty set whose elements have a comparison operation, the set of function values ​​has lower and upper bounds:
.
Here and can be finite numbers or symbols and .

1.1. Let us show that the points and do not belong to the set of function values. That is, a set of values ​​cannot be a segment.

If or is point at infinity: or , then such a point is not an element of the set. Therefore, it cannot belong to multiple values.

Let (or ) be a finite number. Let's assume the opposite. Let the point (or ) belong to the set of function values. That is, there is such for which (or). Let us take points and satisfying the inequalities:
.
Since the function is strictly monotonic, then
, if f increases;
, if f is decreasing.
That is, we have found a point at which the function value is less (greater than ). But this contradicts the definition of the lower (upper) bound, according to which
for everyone.
Therefore, points cannot belong to the set of function values.

1.2. Now we will show that the set of values ​​is an interval, and not a union of intervals and points. That is, for any point there is , for which .

According to the definitions of infimum and supremum, any neighborhood of the points and contains at least one element of the set . Let be an arbitrary number belonging to the interval: . Then for a neighborhood there exists , for which
.
For a neighborhood there is , for which
.

Since and , then . Then
(4.1.1) if it increases;
(4.1.2) if it decreases.
Inequalities (4.1) are easy to prove by contradiction. But you can use, according to which there is an inverse function on the set, which strictly increases if it increases and strictly decreases if it decreases. Then we immediately obtain inequalities (4.1).

So, we have a segment where if increases;
if it decreases.
At the ends of the segment the function takes on the values ​​and . Since, according to the Bolzano-Cauchy theorem, there is a point for which.

Since , then we have thereby shown that for any there is , for which . This means that the value set of the function is an open interval.

2. Now we will show that the inverse function is continuous at an arbitrary point in the interval: . To do this, apply to the segment. Since, the inverse function is continuous on the segment, including at the point.

The theorem has been proven.

Used literature:
O.I. Besov. Lectures on mathematical analysis. Part 1. Moscow, 2004.
CM. Nikolsky. Course of mathematical analysis. Volume 1. Moscow, 1983.

See also:

2.Theory of inverse functions

Inverse trigonometric functions

Definition of an inverse function

Definition. If a function f(x) defines a one-to-one correspondence between its domain X and its domain Y (in other words, if any different values ​​of the argument correspond to different values ​​of the function), then the function f(x) is said to have inverse function or what functionf(x) is reversible.

Definition. The inverse function is a rule that tells each number atє U matches the number Xє X, and y=f(x). Inverse domain

a function is a set Y, the range of values ​​is X.

Root theorem. Let the function f increase (or decrease) on the interval I, the number a is any of the values ​​​​accepted by f on this interval. Then the equation f(x)=a has a single root in the interval I.

Proof. Let us consider an increasing function f (in the case of a decreasing function the reasoning is similar). By condition, in the interval I there exists a number b such that f(b)=a. Let us show that b is the only root of the equation f(x)=a.

Let us assume that there is another number on the interval I c≠ b, such that f(c)=a. Then or with b. But the function f increases on the interval I, therefore, accordingly, either f(c) f(b). This contradicts the equality f(c)= f(b)=a. Consequently, the assumption made is incorrect and in the interval I, except for the number b, there are no other roots of the equation f(x) = a.

Inverse function theorem. If a function f increases (or decreases) on the interval I, then it is invertible. The inverse function g of f, defined in the range of values ​​of f, is also increasing (respectively decreasing).

Proof. For definiteness, let us assume that the function f is increasing. The invertibility of the function f is an obvious consequence of the root theorem. Therefore, it remains to prove that the function g, inverse to f, is increasing on the set E(f).

Let x 1 and x 2 be arbitrary values ​​from E(f), such that x 2 > x 1 and let y 1 = g (x 1), y 2 = g ( x 2 ). By definition of the inverse function, x 1 = f(y 1) and x 2 = f(y 2).

Using the condition that f is an increasing function, we find that the assumption y 1≥ y 2 leads to the conclusion f(y 1) > f(y 2), that is, x 1 > x 2. This

contradicts the assumption x 2 > x 1 Therefore, y 1 > y 2, that is, from the condition x 2 > x 1 it follows that g(x 2)> g(x 1). Q.E.D.

The original function and its inverse are mutually reverse.

Graphs of mutually inverse functions

Theorem. The graphs of mutually inverse functions are symmetrical with respect to the straight line y=x.

Proof. Note that from the graph of the function f one can find the numerical value of the function g inverse to f at an arbitrary point a. To do this, you need to take a point with a coordinate not on the horizontal axis (as is usually done), but on the vertical one. From the definition of the inverse function it follows that the value of g(a) is equal to b.

In order to depict the graph of g in the usual coordinate system, it is necessary to display the graph of f relative to the straight line y=x.

Algorithm for composing the inverse function for the function y=f(x), x X.

1. Make sure that the function y=f(x) is invertible on X.

2. From the equation y=f(x) x express through y, taking into account that x є X .

Z. In the resulting equality, swap x and y.

2.2.Definition, properties and graphs of inverse trigonometrics

functions

arcsine

The sine function increases on the segment
and takes all values ​​from -1 to 1. Therefore, by the root theorem, for any number a such that
, in the interval there is a single root of the equation sin x = a. This number is called the arcsine of the number a and is denoted by arcsin a.

Definition. The arcsine of a number a, where , is a number from a segment whose sine is equal to a.

Properties.

D(y) = [ -1;1 ]

E(y) = [-π/2;π/2]

y (-x) = arcsin(-x) = - arcsin x – the function is odd, the graph is symmetrical about the point O(0;0).

arcsin x = 0 at x = 0.

arcsin x > 0 at x є (0;1]

arcsin x< 0 при х є [-1;0)

y = arcsin x increases for any x є [-1;1]

1 ≤ x 1< х 2 ≤ 1 <=>arcsin x 1< arcsin х 2 – функция возрастающая.

arc cosine

The cosine function decreases on the segment and takes all values ​​from -1 to 1. Therefore, for any number a such that |a|1, on the segment there is a single root in the equation cosx=a. This number b is called the arccosine of the number a and is denoted by arcos a.

Definition . The arc cosine of a number a, where -1 a 1, is a number from the segment whose cosine is equal to a.

Properties.

1. E(y) =

   y(-x) = arccos(-x) = π - arccos x – the function is neither even nor odd.

   arccos x = 0 at x = 1

   arccos x > 0 at x є [-1;1)

arccos x< 0 – нет решений

y = arccos x decreases for any x є [-1;1]

1 ≤ x 1< х 2 ≤ 1 <=>arcsin x 1 ≥ arcsin x 2 – decreasing.

Arctangent

The tangent function increases on the segment -
Therefore, by the root theorem, the equation tgx=a, where a is any real number, has a unique root x on the interval -. This root is called the arctangent of the number a and is denoted arctga.

Definition. Arctangent of the number aR this number is called x , whose tangent is equal to a.

Properties.

E(y) = (-π/2;π/2)

y(-x) = y = arctg(-x) = - arctg x – the function is odd, the graph is symmetrical about the point O(0;0).

arctg x = 0 at x = 0

The function increases for any x є R

-∞ < х 1 < х 2 < +∞ <=>arctg x 1< arctg х 2

Arccotangent

The cotangent function on the interval (0;) decreases and takes all values ​​from R. Therefore, for any number a in the interval (0;) there is a unique root of the equation cotg x = a. This number a is called the arccotangent of the number a and is denoted by arcctg a.

Definition. The arc cotangent of a number a, where a R, is a number from the interval (0;) , whose cotangent is equal to a.

Properties.

E(y) = (0;π)

y(-x) = arcctg(-x) = π - arcctg x – the function is neither even nor odd.

arcctg x = 0– does not exist.

Function y = arcctg x decreases for any x є R

-∞ < х 1 < х 2 < + ∞ <=>arcctg x 1 > arcctg x 2

The function is continuous for any x є R.

2.3 Identical transformations of expressions containing inverse trigonometric functions

Example 1. Simplify the expression:

A) Where

Solution. Let's put
. Then
And
To find
, let's use the relation
We get
But . In this segment, the cosine takes only positive values. Thus,
, that is, where
.

b)

Solution.

Solution. Let's put
. Then
And
Let's find first, for which we use the formula
, where
Since in this interval the cosine takes only positive values, then
.

We have already encountered a problem where, given a given function f and a given value of its argument, it was necessary to calculate the value of the function at this point. But sometimes you have to face the inverse problem: to find, given a known function f and its certain value y, the value of the argument in which the function takes a given value y.

A function that takes each of its values ​​at a single point in its domain of definition is called an invertible function. For example, a linear function would be invertible function. But the quadratic function or the sine function will not be invertible functions. Since a function can take the same value with different arguments.

Inverse function

Let us assume that f is some arbitrary invertible function. Each number from the domain of its values ​​y0 corresponds to only one number from the domain of definition x0, such that f(x0) = y0.

If we now associate each value x0 with a value y0, we will obtain a new function. For example, for a linear function f(x) = k * x + b, the function g(x) = (x - b)/k will be its inverse.

If some function g at every point X range of values ​​of the invertible function f takes a value such that f(y) = x, then we say that the function g- there is an inverse function to f.

If we are given a graph of some invertible function f, then in order to construct a graph of the inverse function, we can use the following statement: the graph of the function f and its inverse function g will be symmetrical with respect to the straight line specified by the equation y = x.

If a function g is the inverse of a function f, then the function g will be an invertible function. And the function f will be the inverse of the function g. It is usually said that two functions f and g are mutually inverse to each other.

The following figure shows graphs of functions f and g mutually inverse to each other.

Let us derive the following theorem: if a function f increases (or decreases) on some interval A, then it is invertible. The inverse function g, defined in the range of values ​​of the function f, is also an increasing (or correspondingly decreasing) function. This theorem is called inverse function theorem.

Lesson objectives:

Educational:

• develop knowledge on a new topic in accordance with the program material;
• study the property of reversibility of a function and teach how to find the inverse function of a given one;

Developmental:

• develop self-control skills, substantive speech;
• master the concept of inverse function and learn methods for finding the inverse function;

Educational: to develop communicative competence.

Equipment: computer, projector, screen, interactive whiteboard SMART Board, handouts (independent work) for group work.

Progress of the lesson.

1. Organizational moment.

Target – preparing students for work in class:

Definition of absentees,

Getting students in the mood for work, organizing attention;

State the topic and purpose of the lesson.

2. Updating students’ basic knowledge. Frontal survey.

Target - establish the correctness and awareness of the studied theoretical material, repetition of the material covered.<Приложение 1 >

A graph of a function is shown on the interactive whiteboard for students. The teacher formulates a task - consider the graph of a function and list the studied properties of the function. Students list the properties of a function in accordance with the research design. The teacher, to the right of the graph of the function, writes down the named properties with a marker on the interactive board.

Function properties:

At the end of the study, the teacher reports that today in the lesson they will become acquainted with another property of a function - reversibility. To meaningfully study new material, the teacher invites the children to get acquainted with the main questions that students must answer at the end of the lesson. The questions are written on a regular board and each student has them as handouts (distributed before the lesson)

1. Which function is called invertible?
2. Is any function invertible?
3. What function is called the inverse of a datum?
4. How are the domain of definition and the set of values ​​of a function and its inverse related?
5. If a function is given analytically, how can one define the inverse function by a formula?
6. If a function is given graphically, how to graph its inverse function?

3. Explanation of new material.

Target - generate knowledge on a new topic in accordance with the program material; study the property of reversibility of a function and teach how to find the inverse function of a given one; develop substantive speech.

The teacher presents the material in accordance with the material in the paragraph. On the interactive whiteboard, the teacher compares the graphs of two functions whose domains of definition and sets of values ​​are the same, but one of the functions is monotonic and the other is not, thereby introducing students to the concept of an invertible function.

The teacher then formulates the definition of an invertible function and conducts a proof of the invertible function theorem using the graph of a monotonic function on the interactive whiteboard.

Definition 1: The function y=f(x), x X is called reversible, if it takes any of its values ​​only at one point of the set X.

Theorem: If a function y=f(x) is monotonic on a set X, then it is invertible.

Proof:

1. Let the function y=f(x) increases by X and let x 1 ≠x 2- two points of the set X.
2. To be specific, let x 1< x 2.
   Then from the fact that x 1< x 2 it follows that f(x 1) < f(x 2).
3. Thus, different values ​​of the argument correspond to different values ​​of the function, i.e. the function is invertible.

(As the proof of the theorem progresses, the teacher uses a marker to make all the necessary explanations on the drawing)

Before formulating the definition of an inverse function, the teacher asks students to determine which of the proposed functions is invertible? The interactive whiteboard shows graphs of functions and writes several analytically defined functions:

B)

G) y = 2x + 5

D) y = -x 2 + 7

The teacher introduces the definition of an inverse function.

Definition 2: Let the invertible function y=f(x) defined on the set X And E(f)=Y. Let's match each one y from Y that's the only meaning X, at which f(x)=y. Then we get a function that is defined on Y, A X– function range

This function is designated x=f -1 (y) and is called the inverse of the function y=f(x).

Students are asked to draw a conclusion about the connection between the domain of definition and the set of values ​​of inverse functions.

To consider the question of how to find the inverse of a given function, the teacher attracted two students. The day before, the children received an assignment from the teacher to independently analyze the analytical and graphical methods of finding the inverse function of a given function. The teacher acted as a consultant in preparing students for the lesson.

Message from the first student.

Note: the monotonicity of the function is sufficient condition for the existence of the inverse function. But it is not a necessary condition.

The student gave examples of various situations when a function is not monotonic but invertible, when a function is not monotonic and not invertible, when it is monotonic and invertible

The student then introduces students to a method for finding the inverse function given analytically.

Finding algorithm

1. Make sure the function is monotonic.
2. Express the variable x in terms of y.
3. Rename variables. Instead of x=f -1 (y) write y=f -1 (x)

Then he solves two examples to find the inverse function of a given one.

Example 1: Show that for the function y=5x-3 there is an inverse function and find its analytical expression.

Solution. The linear function y=5x-3 is defined on R, increases on R, and its range of values ​​is R. This means that the inverse function exists on R. To find its analytical expression, solve the equation y=5x-3 for x; we get This is the required inverse function. It is defined and increasing on R.

Example 2: Show that for the function y=x 2, x≤0 there is an inverse function, and find its analytical expression.

The function is continuous, monotonic in its domain of definition, therefore, it is invertible. Having analyzed the domains of definition and sets of values ​​of the function, a corresponding conclusion is made about the analytical expression for the inverse function.

The second student makes a presentation about graphic method of finding the inverse function. During his explanation, the student uses the capabilities of the interactive whiteboard.

To obtain a graph of the function y=f -1 (x), inverse to the function y=f(x), it is necessary to transform the graph of the function y=f(x) symmetrically with respect to the straight line y=x.

During the explanation on the interactive whiteboard, the following task is performed:

Construct a graph of a function and a graph of its inverse function in the same coordinate system. Write down an analytical expression for the inverse function.

4. Primary consolidation of new material.

Target - establish the correctness and awareness of the understanding of the studied material, identify gaps in the primary understanding of the material, and correct them.

Students are divided into pairs. They are given sheets of tasks, in which they do the work in pairs. The time to complete the work is limited (5-7 minutes). One pair of students works on the computer, the projector turns off during this time and the rest of the children cannot see how the students work on the computer.

At the end of the time (it is assumed that the majority of students have completed the work), the students’ work is shown on the interactive board (the projector is turned on again), where it is determined during the check whether the task was completed correctly in pairs. If necessary, the teacher carries out correctional and explanatory work.

Independent work in pairs<Appendix 2 >

5. Lesson summary. Regarding the questions that were asked before the lecture. Announcement of grades for the lesson.

Homework §10. No. 10.6(a,c) 10.8-10.9(b) 10.12 (b)

Algebra and the beginnings of analysis. Grade 10 In 2 parts for general education institutions (profile level) / A.G. Mordkovich, L.O. Denishcheva, T.A. Koreshkova, etc.; edited by A.G. Mordkovich, M: Mnemosyne, 2007

What is an inverse function? How to find the inverse of a given function?

Definition .

Let the function y=f(x) be defined on the set D, and E be the set of its values. Inverse function with respect to function y=f(x) is a function x=g(y), which is defined on the set E and assigns to each y∈E a value x∈D such that f(x)=y.

Thus, the domain of definition of the function y=f(x) is the domain of values ​​of its inverse function, and the domain of values ​​y=f(x) is the domain of definition of the inverse function.

To find the inverse function of a given function y=f(x), you need :

1) In the function formula, substitute x instead of y, and y instead of x:

2) From the resulting equality, express y through x:

Find the inverse function of the function y=2x-6.

The functions y=2x-6 and y=0.5x+3 are mutually inverse.

The graphs of the direct and inverse functions are symmetrical with respect to the straight line y=x(bisectors of the I and III coordinate quarters).

y=2x-6 and y=0.5x+3 - . The graph of a linear function is . To construct a straight line, take two points.

It is possible to express y unambiguously in terms of x in the case when the equation x=f(y) has a unique solution. This can be done if the function y=f(x) takes each of its values ​​at a single point in its domain of definition (such a function is called reversible).

Theorem (necessary and sufficient condition for the invertibility of a function)

If the function y=f(x) is defined and continuous on a numerical interval, then for the function to be invertible it is necessary and sufficient that f(x) be strictly monotonic.

Moreover, if y=f(x) increases on an interval, then the function inverse to it also increases on this interval; if y=f(x) decreases, then the inverse function decreases.

If the reversibility condition is not satisfied throughout the entire domain of definition, you can select an interval where the function only increases or only decreases, and on this interval find the function inverse to the given one.

A classic example is . In between)