Teaching content: inverse function

Intensive teaching of basic knowledge

1. Basic knowledge chart

2. The concept of inverse function

Let y = f (x) mean that y is a function of the independent variable x, and its domain is a and its domain is c. Solve x by formula y = f (x) and get formula x = φ (y). If for any value of y in C, X has a unique value corresponding to it through x = φ (y), then X.

The definition domain of function y = f (x) is the range of its inverse function y = f- 1 (x); The range of function y = f (x) is the domain of its inverse function y = f- 1 (x).

The image of function y = f (x) and the image of its inverse function y = f- 1 (x) are symmetrical about the straight line y = X. 。

3. Understanding of the concept of inverse function

Inverse function is also a function in essence.

The inverse function is relative to the original function, in other words, the inverse function cannot exist independently of the original function.

Not all functions have inverse functions. For example, the function y = x2 has no inverse function. Only the function with unique original image, that is, it can be inferred that f(x 1)≠f(x2) has an inverse function for any function f(x) where x 1 and x2 is f).

If the function y = f (x) has the inverse function y = f- 1 (x), then the function y = f (x) is also the inverse function of its inverse function y = f- 1 (x), that is, they are reciprocal functions.

The definition domain and value domain of function y = f (x) are the value domain and value domain of its inverse function y = f- 1 (x) respectively.

The domain and value domain of the inverse function should be exactly the domain and the domain of the original function. For example, the function y = (x ∈ z) is not the inverse function of the function y = 2x (x ∈ z), because the domain of the former is obviously not the domain of the latter. Therefore, find the inverse function y = f-65438 of the function y = f (x).

4. To find the inverse function of the function y = f (x) of a given analytical formula, the steps are as follows:

(1) Solve x = f-1(y) from equation y = f (x);

(2) exchange x and y to get y = f-1(x);

(3) Write the definition domain of y = f- 1 (x) according to the value domain of y = f (x).

Two functions that are reciprocal functions, if they have analytic expressions, are generally different, but they are the same. For example, the inverse function of function y = x is still y = x, and the inverse function of function y = is still y =.

5. The relationship between reciprocal function images

In the same rectangular coordinate system, the images of function y = f (x) and its inverse function y = f- 1 (x) are symmetrical about the straight line y = X, especially when the function and its inverse function are the same, the images of the function itself are symmetrical about the straight line y = X. 。

In y = f (x) and x = f- 1 (y), x and y represent the same quantity, but their positions are different. In y = f (x), x is the independent variable and y is the function y = f (x; In x = f- 1 (y), y is an independent variable and x is a function of y. In the same rectangular coordinate system, the images of y = f (x) and x = f- 1 (y) are the same point set.

6. Other properties of inverse function

In Y = f (x) and Y = f- 1 (x), the positions of x and y are the same, but the meanings of the quantities expressed are different.

If y = f (x) (x ∈ a) and y = f- 1 (x) (x ∈ c) are reciprocal functions, then there are

f〔f- 1(x)〕= x(x∈C)；

f- 1〔f(x)〕=x(x∈A)。

Two reciprocal functions have the same monotonicity in their respective domains.

If odd function has an inverse function, its inverse function is also odd function.

Monotone function must have inverse function.

If two mutually inverse images have intersection points, their intersection points are not necessarily on the straight line y = X.

Analysis of key and difficult points

1. Of the three steps to find the inverse function, remember that the third step is essential, that is, determine the domain of the inverse function from the range of the original function y = f (x), and after finding the inverse function, you must give the domain of the inverse function.

2.x = f (y) and y = f- 1 (x) are the same function.

This is because their definition domain and value domain correspond to the same (the value domain and definer of the original function respectively), and the corresponding rules are the same.

3. A method to determine whether the function y = f (x) defined on a has an inverse function.

Let x 1, x2∈A, x 1≠x2, and judge whether f(x 1)≠f(x2) is true. If it is, f(x) has an inverse function on a; If not, then f(x) has no inverse function on a; If a function is monotonous in an interval, then it has an inverse function in this interval.

4. Solve the inverse function of piecewise function

Set piecewise function

Y = has an inverse function. Its inverse function must be solved by segments.

That is, y =

Example 1 Find the inverse function of the following function: (1) y = 3x+4 (x ≤ 0);

(2)y= (- 1≤x≤0)

Solution: (1) from y = 3x+4, x =；;

Both sides are cubes, x2 = () 3.

X has a unique solution on R- if and only if () 3≥0, that is, y≥4.

x=-()。

Exchange x and y to get y =-() (x ≥ 4).

This is the inverse function.

(2) From y =, x2 = 1-Y2 ① is obtained.

If and only if 0≤ 1-y2≤ 1(y≥0), ① has a unique solution on [- 1, 0], that is, x =-.

Exchange x and y to get y =-() (x ∈ [0, 1]).

This is the inverse function.

Comment on the discussion about finding the range of function in ξ 1.6. We introduce the inverse method, that is, the condition of finding the solution of X in the defined range. Here, we look for the conditions that make X have a unique solution in the definition domain. Can you tell us why?

Example 2 It is known that the image with f (x) = and function y = g (x) and the image with function y = f- 1 (x+ 1) are about a straight line y=g(x), then g( 1 1) is equal to f(x)=.

A.B. C. D。

Solution: First find the inverse function of f (x) = (x ≠ 1).

From y =, x = (y ≠ 2).

Exchange x with y to get the inverse function f- 1 (x) = (x ≠ 2) of f(x).

∴f- 1(x+ 1)=。

∵f- 1(x+ 1) and g(x) are symmetric about y = x,

∴f- 1(x+ 1) and g(x) are reciprocal functions.

Order = 1 1, and the solution is x =, ∴ g (1 1) =. So choose B.

The analysis of f- 1(x+ 1) means that the independent variable in the inverse function is replaced by x+ 1, that is, f- 1(x+ 1 instead of x.f-1(x+655)

Example 3 f (x) =, find f- 1 [f (x)] and f- 1 (x)].

Solution: Let y = (x= (y≠2- 1), then x= (y≠2).

∴f- 1(x)= (x≠2)、

f- 1〔f(x)〕= = x(x≦- 1)，

f〔f- 1(x)〕= =x (x≠2)。

The analytic f- 1 [f (x)] and f [f- 1 (x)] are different functions due to different domains, where x∈A and f [f] are in f- 1 [f (x)].

Example 4 Find the inverse of the function f (x) =.

Through analysis, the inverse functions of y = x2- 1 (x ≥ 0) and y = 2x- 1 (x < 0) are obtained, and then written in piecewise form.

Solution: 1 from y = x2- 1, x2 = y+ 1.

If and only if y+ 1≥0, that is, y≥- 1, x has a unique solution on [0, +∞], that is, x =.

So the inverse function of y = x2- 1 (x ≥ 0) is y = (x ≥- 1).

2 from y = 2x- 1, x = ①.

∵ x < 0, i.e. < 0, y

If and only if y

Therefore, the inverse function of y = 2x- 1 (x < 0) is y = (x

From 1 2, the inverse function is

f- 1(x)= 0

Unsolvable ingenious inspiration

Example 1 It is known that the image of the function f(x)=(a≦) is symmetrical about the straight line y = x, and the value of a is found.

It is analyzed that the so-called function image is symmetrical about the straight line y = x, that is, this function and its inverse function are the same function.

Solution: y = (x= (y≠2-a), x= (y≠2).

∴f- 1(x)= (x≠2)。

The image of the function f(x) is symmetrical about the straight line y = x,

∴f(x) and f- 1(x) are the same function.

∴-a=2，

∴a=-2.

If two functions are the same, then their corresponding rules are the same, and their domains are the same.

For the real number A that is neither 0 nor 1, the image of the function Y = is always symmetrical about the straight line y = X. Can you prove this conclusion?

Example 2 It is known that the function y = f(x) has a definition domain and a value domain of c, and the inverse function f- 1(x) exists. If f (x) is increasing function on a, it is proved that f- 1(x) is increasing function on C. 。

This analysis is proved according to the monotonicity definition of the function.

Certificate: let x 1, x2∈C, and X 1

x 1=f(y 1)，x2=f(y2)。

∴f(y 1)<； f(y2)，

And f (x) is the increasing function of a,

∴y 1<； y2，

That is f-1(x1) < f-1(x2),

Therefore, f- 1(x) is the increasing function on C. 。

The evaluation function f(x) and its inverse function f- 1(x) have the same increase or decrease.

Example 3 It is known that the function Y = AX+B (A ≠ 0) has an inverse function, and its inverse function is itself, so the conditions that real numbers A and B should meet.

Analysis shows that if the point (x0, y0) is on the original function image, then the point (y0, x0) is also on the function image.

Solution: If point (x0, y0) is any point on the image of function y = ax+b (a ≠ 0), then point (y0, x0) is the point on the image of its inverse function. Because the inverse function of the original function is itself, (y0, x0) is also in the function y = ax+b (a).

∴

∴ (a2- 1)x0+b(a+ 1)=0，

∴ A2- 1 = 0, and b (A+ 1) = 0,

∴ or

Example 3 and example 1 are the same type of problems, but different solutions are given. Please taste it carefully.

Textbook problem solving

Page 69 of the textbook, exercise 2.4, answer to question 4.

The inverse of the function y = x+b is y = 5x-5b.

Suppose y = ax+3 is the inverse function of y = x+b,

Therefore, function y = 5x-5b and function y = ax+3 are the same function, which leads to

solve

Answer to question 5:

Prove: Find the inverse function of function y =(x≦- 1).

∫x≦- 1( 1+x)y = 1-x

∴x( 1+y)= 1-y

If y≦- 1, x =, x and y are interchanged:

y =(x≦- 1)

It is proved that the inverse function of function y =(x≦- 1) is the function itself.

The image of the function (1) is symmetrical about the straight line y = X. Using this feature, we can know that if there is a little P(a, b) on the image, there must be a little P'(b, a).

② Because the function can be transformed into y =- 1(x≦- 1).

Therefore, the image of the inverse proportional function can be obtained by shifting it to the right and down by one unit.

Question 6 (1): Y = f (x) = 2x+3 is the same as X = f- 1 (y) = (y-3).

In the same coordinate system, the image of y = f (x) is the same as that of y=f(x = f- 1 (y).

(2) Example: f (x) = x3, f- 1 (x) = Two function images are symmetrical about the straight line y = x. 。

In the same coordinate system, the images of Y = f (x) and its inverse function Y = f- 1 (x) are symmetrical about the straight line y = X. 。

Propositional trend analysis

(1) This knowledge is mainly examined in the college entrance examination: ① According to the analytical formula of the original function, the analytical formula of the inverse function can be obtained; ② Using the relationship between the original function and the inverse function image to solve the problem; (3) Solving problems by using the relationship between the domain and the range of the original function and the inverse function; ④ Draw an image to solve the image problem.

(2) The previous college entrance examination questions were mainly multiple-choice questions and fill-in-the-blank questions, most of which were middle and low-grade questions, focusing on concepts.

(3) In the process of thinking and solving problems, we mainly use the idea of equation and the thinking method of combining numbers and shapes.

(4) The concept of inverse function frequently appears in college entrance examination questions, such as the symbol and meaning of inverse function, the method of finding inverse function and the relationship between images as inverse function. In particular, finding the inverse function is the most important. The concept and image of inverse function are combined with quadratic function, exponential function and logarithmic function. It is still the direction of college entrance examination in the future to investigate the definition, range, image, monotonicity and parity of inverse function.

Typical hot issues

Example 1 1999 math (literature and history) problem (9) in the college entrance examination, given the function y = (x ∈ r, and x≠ 1), its inverse function is ().

A.y = (x ∈ r and x≠ 1) b.y = (x ∈ r and x≠6).

C.y = (x ∈ r and x≦-)d. y =(x∈r and x≦-5).

It differs from the textbook examples only in the coefficient of molecules.

In the textbook, the steps of finding the inverse function of a function are illustrated as follows: (1) x = f-1(y) from y = f (x); (2) exchange x and y and rewrite it as y = f-1(x); (3) Determine the domain of the inverse function from the range of y = f (x).

Example 2 Let the function y = 1-(- 1 ≤ x ≤ 0), then the image of the function y = f- 1 (x) is () in the following figure.

Solution 1:

∵y = 1- is defined as [- 1, 0], ∴ its inverse function y = f- 1 (x) is defined as [- 1, 0], and A and C can be excluded.

How to choose b and d? According to the basic property that two mutually inverse images are symmetrical about a straight line Y = X, we can choose a special point.

∫ y = f (x) = 1- (-1≤ x ≤ 0) is the passing point (-,1-).

Image ∴ y = f- 1 (x) must pass (1-,-).

In d, when y =-, x is close to 1, so X ≠ 1-< 0. 15.

You should choose B.

Solution 2: f (x) = 1-(- 1 ≤ x ≤ 0)

Its value range is y ∈ [0, 1], and the image of the original function is known as A. According to the symmetry of the image of the original function and its inverse function about the straight line y = x, it is known that the image of f- 1(x) should be B.

You should choose B.

Note that this topic mainly examines the related concepts of inverse function, requires a deep understanding of the relationship between the original function and inverse function, and also examines the idea of combining numbers with shapes.

Example 3 If the inverse function of the function y = f (x) is y = g (x), f (a) = b and ab ≠ 0, then g(b) is equal to ().

a a b a- 1 c b d b- 1

The analysis of this topic mainly investigates the nature and application of inverse function. According to the topic, it can also be verified by special functions and special points.

Solution: ∵ Point (a, b) is on the image of the original function y = f (x),

∴(b,a) should be on the image of its inverse function y = f- 1 (x).

∴g(b)=a.

You should choose a.

Example 4 The inverse function of the function f (x) = (x- 1)+2 is f- 1 (x) =.

Solution: y = (x- 1)+2, (x- 1) = y-2, x- 1 = (y-2) 3, x = (y-2) 3+ 1, so

∴ (x-2)3+ 1 should be filled in.

Knowledge verification experiment

1. By studying students' learning behavior, psychologists find that students' acceptance depends on the time it takes teachers to introduce concepts and describe problems. At the beginning of the lecture, students' interest surged; For a short time, students' interest remained in an ideal state, and then students' attention began to disperse. The analysis results and experiments show that f(x) represents the students' ability to master and accept concepts, and x represents the time (unit: minutes) for putting forward and teaching concepts, and the following formula can be adopted:

f(x)= 1

(1) How many minutes after the lecture, are the students most receptive? How long can it last?

(2) When is the students' receptive ability stronger than 20 minutes after the lecture?

(3) A math problem needs 55% acceptance, 13 minutes. Can the teacher finish teaching this question in time when the students reach the required acceptance?

(4) If students' acceptance is measured every 5 minutes, and then the average m = is calculated, can it be higher than 45?

Solution: (1) When 0 < x ≤ 10,

f(x)=-0. 1x 2+2.6x+43 =-0. 1(x- 13)2+59.9，

So f(x) is increasing, and the maximum value is f (10) =-0.1× (-3) 2+59.9 = 59; Obviously, when 16 < x ≤ 30, f(x) decreases, and F (x)

Therefore, after the lecture 10 minutes, students reach the strongest acceptance (value 59) and keep it for 6 minutes.

(2)f(5)=-0. 1×(5- 13)2+59.9 = 59.9-6.4 = 53.5

F(20)=-3×20+ 107=470) can be washed once, or it can be washed twice after the grade is divided into two parts. Which scheme is used to wash vegetables with less pesticide residues? Explain why.

Solution: (1) f (0) = 1 means that the amount of pesticides on vegetables remains unchanged without washing with water.

(2) The conditions and specific properties of the function f(x) are f (0) = 1, and f (1) =.

On [0, +∞], f(x) decreases monotonically, and 0 < f (x) ≤ 1.

(3) Suppose that the pesticide residue after only one cleaning is F 1 = the pesticide residue after two cleaning is F2 = =.

Then f 1-F2 =-=.

Therefore, when a > 2, f1> F2;

When a = 2, f1= F2;

When 0 < a < 2, f 1 < F2.

Therefore, when a>2 points, the pesticide residue is less after washing twice;

When a = 2, the two cleaning methods have the same effect;

When 0 < a < 2, the pesticide residue in one cleaning is less.

Synchronous outline exercise

First, multiple choice questions

Inverse function of 1 Y = a-(x ≥ a) is ()

a . y =(x-a)2+a(x≥a)b . y =(x-a)2-a(x≥a)

c . y =(x-a)2+a(x≤a)d . y =(x-a)2-a(x≤a)

2. It is known that the function y = f (x) has an inverse function, so the root of the equation f (x) = 0 is ().

A. there is only one real root. B. There is at most one real root.

C. there is at least one real root D.0, 1 or greater than 1.

3. If the point (a, b) is on the image of y = f (x), then the point on the inverse function image of the later point must be ().

A.(a，f- 1(a)) B.(f- 1(b)，b) C.(f- 1(a)，a) D.(b，f- 1(b))

4. There are three functions, the first function is y = f (x), its inverse function is the second function, and the images of the third function and the second function are symmetrical about the origin, so the third function is ().

a . y =-f(x)b . y = f- 1(-x)c . y =-f- 1(-x)d . y = f- 1(x)

5. The image of function y = f (x) passes through the third and fourth quadrants, so the image of function y =-f- 1 (x) passes through ().

A. First and second quadrants B. Second and third quadrants C. Third and fourth quadrants D. First and fourth quadrants

6. Among the following intervals, the interval where y = 2 | x | has no inverse function is ().

A.〔2，4〕b .〔3〕-4，4〕 C.〔0，+∞〕 D.(-∞，0)

7. If the image of function y = f- 1 (x) passes through point (-2,0), then the image of function y = f (x+5) passes through point ().

A.(5，-2) B.(-2，-5) C.(-5，-2) D.(2，-5)

Second, fill in the blanks

The range of 1. function y = is.

2. Given that the function f(x) is defined on (-∞, 0] and f (x+ 1) = x2+2x, then f- 1 (1) =.

3. If the line Y = AX+2 and the line Y = 3x-b are symmetrical about the line Y = X, then a =, b =.

4. If the image of function f(x)=(a≦) is symmetric about y = x, then a =.

5. The image of the inverse function of the function f (x) = AX3+AX- 1

6. Given that the inverse function of f (x) = is itself, then a =, b =.

7. Does 7.Y = have an inverse function? ; When x ∈ [0,], the inverse function is and the domain is; When x ∈ [-, 0], the inverse function is and the domain is.

8. It is known that f (x) = (x ∈ r and x≦-), and the value of f- 1(2) is.

Third, answer questions.

1. Does the function f (x) = x-n (x < 0, n ∈ z) have an inverse function? If there is no explanation, if there is, find f- 1(x) and judge whether it is a increasing function or a subtraction function.

2. It is known that F(x) = x2 and g (x) = x+5. Let f (x) = f [g-1(x)]-g-1[f (x)]. Try to find the minimum value of f (x).

3. It is known that the inverse function of the function y = f (x) is y = f- 1 (x).

(1) Try to find the inverse function of function y = f (MX+N) (m ≠ 0);

(2) Try to find the inverse function of function y = f (ax3+b) (a ≠ 0).

Quality optimization training

1. Find the inverse of the function f (x) =.

2. Let the function f (x) =, the image of the known function y = g (x) and the image of y = f- 1 (x+ 1) are symmetrical about the straight line y=g(x), and find the value of g(3).

3. It is known that f(x)=(x≦-a, a≦).

(1) Find the inverse function of f(x);

(2) If f (x) = f- 1 (x), find the value of a;

(3) How to make the image with y = f- 1 (x) meet the condition in (2).

Reference answer:

Synchronous outline exercise

I.1.c2.b3.d4.c5. Select B6.b7.c

Second, 1. {y | y ∈ r and y≦-} 2. -3.a = b = 64。 a =-55。 (0,-1) 6.0, nonzero real number 7. No; y =； 〔0,4〕; y =-； 〔0,4〕 8.-

Third, when 1. N = 0, f (x) = 1, no inverse function.

When n is non-zero even number, f-1(x) =-=-x (x >; 0)①n & gt； 0,

And n∈Z, f- 1(x) is increasing function, ② n.

When n is odd, y = x-n (x

Inverse function f- 1 (x) = x (x

②n & lt； 0 and n∈Z, f- 1(x) are increasing function 2. -90.

3.( 1)y= f- 1(x)- (2)y=

Quality optimization training

1 . f(x)= 1

2.

3. Solution: (1) y = (x ≠ 2)

(2)a=-2

(3) f- 1 (x) = = 2+(x ≠ 2y ≠ 2)。 To get the image of y = f- 1 (x), just move y = 2 units to the right, and then move up 2 units, and you will get y = f-6550.