§7. Inverse function and its graph

Lesson Objectives:

Educational:

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

Developing:

develop self-control skills, subject speech;
master the concept of an inverse function and learn the methods of finding an inverse function;

Educational: to form communicative competence.

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

During the classes.

1. Organizational moment.

Target – preparing students for work in the classroom:

Definition of absent,

Attitude of students to work, organization of attention;

Message about the topic and purpose of the lesson.

2. Updating the basic knowledge of students. front poll.

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

A graph of the function is shown on the interactive whiteboard for students. The teacher formulates the task - to consider the graph of the function and list the studied properties of the function. Students list the properties of a function according to 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 whiteboard.

Function properties:

At the end of the study, the teacher reports that today at the lesson they will get acquainted with one more property of the function - reversibility. For a meaningful study of new material, the teacher invites the children to get acquainted with the main questions that students must answer at the end of the lesson. Questions are written on an ordinary board and each student has a handout (distributed before the lesson)

What is a reversible function?
Is every function reversible?
What is the inverse given function?
How are the domain of definition and the set of values of a function and its inverse function related?
If the function is given analytically, how do you define the inverse function with a formula?
If a function is given graphically, how to plot its inverse function?

3. Explanation of new material.

Target - to form knowledge on a new topic in accordance with the program material; to study the property of the invertibility of a function and to teach how to find a function inverse to a given one; develop subject matter.

The teacher conducts a presentation of the material in accordance with the material of the paragraph. On the interactive board, 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 bringing students under 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 the 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 the function y=f(x) is monotone on the set X , then it is invertible.

Proof:

Let the function y=f(x) increases by X let it go x 1 ≠ x 2- two points of the set X.
For definiteness, let x 1< x 2.
Then from what x 1< x 2 follows that f(x 1) < f(x 2).
Thus, different values of the argument correspond to different values of the function, i.e. the function is reversible.

(During the proof of the theorem, the teacher makes all the necessary explanations on the drawing with a marker)

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

G) y = 2x + 5

D) y = -x 2 + 7

The teacher introduces the definition of an inverse function.

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

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

Students are invited to draw a conclusion about the relationship between the domain of definition and the set of values of inverse functions.

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

Message from the first student.

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

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

Then the student introduces students to the method of finding the inverse function given analytically.

Finding algorithm

Make sure the function is monotonic.
Express x in terms of y.
Rename variables. Instead of x \u003d f -1 (y) they write y \u003d f -1 (x)

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

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

Decision. The linear function y=5x-3 is defined on R, increases on R, and its range is R. Hence, the inverse function exists on R. To find its analytical expression, we solve the equation y=5x-3 with respect to x; we get This is the desired inverse function. It is defined and increases by R.

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

The function is continuous, monotone in its domain of definition, therefore, it is invertible. Having analyzed the domains of definition and the set 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 how to find the inverse function. In the course of his explanation, the student uses the capabilities of the interactive whiteboard.

To get the 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 fixation of the new material.

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

Students are divided into pairs. They are given sheets with tasks in which they work in pairs. Time to complete the work is limited (5-7 minutes). One pair of students works on the computer, the projector is turned off for 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 completed the work), the interactive whiteboard (the projector turns on again) shows the work of the students, where it is clarified during the test that the task was completed in pairs. If necessary, the teacher conducts corrective, explanatory work.

Independent work in pairs<Appendix 2 >

5. The result of the lesson. On the questions that were asked before the lecture. Announcement of grades for the lesson.

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

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

Suppose we have some function y = f (x) that is strictly monotonic (decreasing or increasing) and continuous on the domain x ∈ a ; b; its range of values is y ∈ c ; d , and on the interval c ; d at the same time, we will have a function x = g (y) with a range of values a ; b. The second function will also be continuous and strictly monotonic. With respect to y = f (x) it will be an inverse function. That is, we can talk about the inverse function x = g (y) when y = f (x) will either decrease or increase on a given interval.

These two functions, f and g , will be mutually inverse.

Why do we need the concept of inverse functions at all?

We need this to solve the equations y = f (x) , which are written just using these expressions.

Let's say we need to find a solution to the equation cos (x) = 1 3 . All points will be its solutions: x = ± a rc c o s 1 3 + 2 π k , k ∈ Z

Inverse with respect to each other will be, for example, the arccosine and cosine functions.

Let us analyze several problems for finding functions inverse to given ones.

Example 1

Condition: what is the inverse function for y = 3 x + 2 ?

Decision

The domain of definitions and the domain of values of the function specified in the condition is the set of all real numbers. Let's try to solve this equation through x, that is, by expressing x through y.

We get x = 1 3 y - 2 3 . This is the inverse function we need, but here y will be an argument, and x will be a function. Let's rearrange them to get a more familiar notation:

Answer: the function y = 1 3 x - 2 3 will be inverse for y = 3 x + 2 .

Both mutually inverse functions can be plotted as follows:

We see the symmetry of both graphs with respect to y = x . This line is the bisector of the first and third quadrants. We have obtained a proof of one of the properties of mutually inverse functions, which we will discuss later.

Let's take an example in which you need to find the logarithmic function, the inverse of a given exponential.

Example 2

Condition: determine which function will be inverse for y = 2 x .

Decision

For a given function, the domain of definition is all real numbers. The range of values lies in the interval 0 ; +∞ . Now we need to express x through y, that is, solve the indicated equation through x. We get x = log 2 y . Rearrange the variables and get y = log 2 x .

As a result, we have obtained exponential and logarithmic functions, which will be mutually inverse to each other over the entire domain of definition.

Answer: y = log 2 x .

On the graph, both functions will look like this:

Basic properties of mutually inverse functions

In this subsection, we list the main properties of functions y = f (x) and x = g (y) that are mutually inverse.

Definition 1

We already derived the first property earlier: y = f (g (y)) and x = g (f (x)) .
The second property follows from the first: the domain of definition y = f (x) will coincide with the domain of the inverse function x = g (y) , and vice versa.
The graphs of functions that are inverse will be symmetrical with respect to y = x .
If y = f (x) is increasing, then x = g (y) will also increase, and if y = f (x) is decreasing, then x = g (y) will also decrease.

We advise you to carefully consider the concepts of the domain of definition and scope of functions and never confuse them. Let's say that we have two mutually inverse functions y = f (x) = a x and x = g (y) = log a y . According to the first property, y = f (g (y)) = a log a y . This equality will be true only in case of positive values of y , and for negative values, the logarithm is not defined, so do not rush to write down that a log a y = y . Be sure to check and add that this is only true for positive y .

But the equality x \u003d f (g (x)) \u003d log a a x \u003d x will be true for any real values of x.

Do not forget about this point, especially if you have to work with trigonometric and inverse trigonometric functions. So, a r c sin sin 7 π 3 ≠ 7 π 3 because the range of the arcsine is π 2 ; π 2 and 7 π 3 are not included in it. The correct entry will be

a r c sin sin 7 π 3 \u003d a r c sin sin 2 π + π 3 \u003d \u003d \u003d in the form of a s u l p r i o n i o n \u003d a r c sin sin π 3 \u003d π 3

But sin a r c sin 1 3 \u003d 1 3 is the correct equality, i.e. sin (a r c sin x) = x for x ∈ - 1 ; 1 and a r c sin (sin x) = x for x ∈ - π 2 ; π 2 . Always be careful with the scope and scope of inverse functions!

Basic mutually inverse functions: power

If we have a power function y = x a , then for x > 0 the power function x = y 1 a will also be inverse to it. Let's replace the letters and get y = x a and x = y 1 a respectively.

On the chart, they will look like this (cases with positive and negative coefficient a):

Basic mutually inverse functions: exponential and logarithmic

Let's take a, which will be a positive number, not equal to 1 .

Graphs for functions with a > 1 and a< 1 будут выглядеть так:

Basic mutually inverse functions: trigonometric and inverse trigonometric

If we need to plot the main branch of the sine and arcsine, it will look like this (shown in the highlighted light area).

Definition of an inverse function and its properties: lemma on the mutual monotonicity of 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 strictly monotonic on a segment, interval, and half-interval. Examples of inverse functions. An example of a problem solution. Proofs of properties and theorems.

Content

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

Definition and properties

Definition of the inverse function
Let the function have a domain X and a set of values Y . And let it have the property:
for all .
Then for any element from the set Y, only one element of the set X can be associated, for which . This correspondence defines a function called inverse function to . The inverse function is denoted as follows:
.

It follows from the definition that
;
for all ;
for all .

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

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

For an increasing function . For descending - .

Theorem on the existence and continuity of the 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 is strictly increasing (decreasing).

For an increasing function .
For descending: .

In a similar way, one can formulate a theorem on the existence and continuity of an 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 it is 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 proved in the same way as the theorem on the existence and continuity of the inverse function on an interval.

Examples of inverse functions

Arcsine

Plots 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 the domain of definition is narrowed, then monotonous sections can be distinguished. So, on the segment , the function is defined, continuous, strictly increasing and taking values from -1 before +1 . Therefore, it has an inverse function on it, which is called the arcsine. The arcsine has a domain of definition and a set of values.

Logarithm

Plots 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 . The set of its values is an open interval. The inverse function is the base two logarithm. It has a scope and a set of values.

Square root

Plots y=x 2 and inverse function.

The power function is defined and continuous for all . The set of its values is a half-interval. But it is not monotonic for all values of the argument. However, on the half-interval it is continuous and strictly monotonically increasing. Therefore, if, as a domain, we take the set, then there is an inverse function, which is called the square root. The inverse function has a domain of definition 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 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 us apply the arithmetic properties of the limits of the function . Since , then only the first term is nonzero:
.
Continuity has been proven.

Let us prove that the function (P1) strictly increases as .
Let's take arbitrary numbers connected by inequalities:
, , .
We need to show that . Let's introduce variables. Then . Since , it is seen from (A2) that . Or
.
Strict increase is proved.

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

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

Proofs of properties and theorems

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

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

Let's assume the opposite. Let there be numbers , so . Let at the same time. Otherwise, we change the notation so that it is . Then, due to the strict monotonicity of f , one of the inequalities must hold:
if f is strictly increasing;
if f is strictly decreasing.
I.e . There was a contradiction. 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 notation:
. That is, we need to prove that if , then .

Let's assume the opposite. Let , but .

If , then . This case is out.

Let be . Then, due to the strict increase of the function , , or . There was a contradiction. Therefore, only the case is possible.

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

Proof of a property on the symmetry of graphs of direct and inverse functions

Let be an arbitrary point of the direct function graph:
(2.1) .
Let's show that the point , symmetrical to the point A with respect to the line , belongs to the graph of the inverse function :
.
It follows from the definition of the inverse function 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 drop perpendiculars on the coordinate axes. Then
, .

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

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

Let's consider a triangle. Since , then
.
The same applies to the triangle:
.
Then
.

Now we find:
;
.

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

Since we have chosen point A arbitrarily, this applies to all points of the graph:
all points of the graph of the function, reflected symmetrically with respect to the straight line, belong to the graph of the inverse function.
Then we can swap places. As a result, we get
all points of the graph of the function, reflected symmetrically about the straight line, belong to the graph of the function.
It follows that the graphs of the 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 denotes the domain of definition of the function - the segment .

1. Let's show that the set of function values is the interval :
,
where .

Indeed, since the function is continuous on the segment , then, according to the Weierstrass theorem, it reaches its minimum and maximum on it. Then, according to the Bolzano-Cauchy theorem, the function takes all values from the segment. That is, for any 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 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 interior point of the segment : . Let us prove that the inverse function is continuous at this point.

Let it correspond to the point . 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 all .

Note that we can take arbitrarily small. Indeed, if we have found a function such that inequalities (3.1) are satisfied for sufficiently small values of , then they will automatically be satisfied for any large values of , if we set for .

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

.

We transform the first inequality (3.1):
(3.1) for all .
;
;
;
(3.2) .
Since it is strictly monotonic, it follows that
(3.3.1) , if 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 exists δ, so |f -1 (y) - f -1 (y 0) |< ε for all |y - y 0 | < δ .

Inequalities (3.3) define an open interval whose ends are separated from the point by distances and . Let there be the smallest of these distances:
.
Due to the strict monotonicity of , , . That's why . 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 have found that for sufficiently small , exists , so that
at .
Now let's change the notation.
For small enough , there exists such that
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 arguments remain the same. Only one-sided neighborhoods of these points need to be considered. 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 , because for any sufficiently small there is , so that
at .

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

Let denotes the domain 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's 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 values of the function. That is, the 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 a set of values.

Let (or ) be a finite number. Let's assume the opposite. Let the point (or ) belong to the set of values of the function . That is, there exists such for which (or ). 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 where the value of the function is less (greater than ). But this contradicts the definition of the lower (upper) face, according to which
for all .
Therefore, the points and cannot belong to the set of values of the function .

1.2. Now let's show that the set of values is an interval and not a union of intervals and points. That is, for any point exists , for which .

According to the definitions of the lower and upper bounds, 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 the neighborhood there exists for which
.
For a neighborhood, there exists for which
.

Since and , then . Then
(4.1.1) if 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 the values and . Since , then by the Bolzano - Cauchy theorem , there exists a point for which .

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

2. Now let's show that the inverse function is continuous at an arbitrary point of the interval : . To do this, apply to the segment. Since , then the inverse function is continuous on the interval , including at the point .

The theorem has been proven.

References:
O.I. Demons. Lectures on mathematical analysis. Part 1. Moscow, 2004.
CM. Nikolsky. Course of mathematical analysis. Volume 1. Moscow, 1983.