Why are the inverse and primitive value ranges reversed from the definition domains?

The image of a primary function is a skewed straight line.

Defining a domain can start from negative infinity and go up to positive infinity.

The same goes for value ranges.

So why are the definition domains and value ranges of a primary function r?

In general, let the domain of the function y=f(x)(x a) be c, and according to the relationship between x and y in this function, x is represented by y, and x= is obtained

f(y).If for any value of y in c, through x=f(y), x has a unique value in a corresponding to it, then x=f(y) means that y is an independent variable and x is a function of the dependent variable y, and such a function x=f(y)(y c) is called the inverse of the function y=f(x)(x a), denoted as y=f -1(x)

Because to say that this is the definition of an inverse function.

There is no why.

It's like an axiom.

Here's how to find it:Let the original function y=ax+b

Formation x=(y-b) a

Then write y=(x-b) a

It's its inverse function.

Let the original function y=x +b

x= (y-b) (y-b 0).

Then write y= (x-b) (x-b 0).

It's its inverse function.

After finding the domain, pay attention to the definition domain and the value range, the definition domain of the inverse function is the value domain of the original function, and the value range of the inverse function is the definition domain of the original function.

Define the representation of the domain:There are three ways to define domain representation: inequality, interval, and set.

The domain of y=[ (3-x)] [lg(x-1)] can be expressed as: 1) x 1;2）x∈(-1]；3）。

Let a, b be two non-empty sets of numbers, and if, according to a certain correspondence f, there is a uniquely definite number f(x) corresponding to any number x in set a, then f:a--b is a function from set a to set b, denoted as y=f(x), and x belongs to set a. where x is called the independent variable, and the value range of x a is called the definition domain of the function.

Find a monotonic interval, which is the domain of the annoying function.

Think of a function as an equation: y=f(x).

Solve the equation to find the expression of x identified by y, x=f (-1)(y) Swap x,y to get the inverse function expression: y=f (-1)(x) For example: find the inverse function of y=3x+5, the function is monotonic in (-, and the range is:

So the domain of the inverse function is: (-The range is: (-is solved by y=3x+5:

x=1 3*y-5 3 The inverse function is: y=1 3*x-5 3 x (-e.g. y=x 2, x=positive and negative root number y, then the inverse function of f(x) is the positive and negative root number x, after finding it, pay attention to the definition domain and value range, the definition domain of the inverse function is the value range of the original function, and the value range of the inverse function is the definition domain of the original function.

The definition domain of the inverse function is the value range of the original function.

Therefore, as long as the value range of the original function is found in the definition domain of the original function, that is, the definition domain of the inverse function.

The inverse function definition domain is the same as the original function value domain.

The inverse function domain is the same as the original function definition domain, and in general, let the domain of the function y=f(x)(x a) be c, according to which x,y in this function

, denoted by y, gives x=

g(y).If for any value of y in c, through x=g(y), x has a unique value corresponding to it in a, then x=g(y) means that y is an independent variable and x is a function of the dependent variable y, and such a function x=g(y)(y c) is called the inverse function of the function y=f(x)(x a), denoted as y=f -1(x)

The domain and range of the inverse function y=f -1(x) are the domain and domain of the function y=f(x), respectively.

Yes, the domains of the original and inverse functions are interchangeable with the range of values.

In other words: the domain of the inverse function must be the domain of the original function;

The domain of the original function must be the domain of the inverse function.

When it is difficult to define the domain of the original function, we can convert it to the range of the inverse function.

This is the domain method for finding a defined domain.

Vice versa. It is reasonable to say that it doesn't matter what happens, and the domains of the two must be equal.

For example, the definition domain of arctanx 1 is:Define the fields 2 x -2 and x ≠0.

Solution ideas: 1. Look at 1 x, the denominator is not 0, so x ≠ 0

2. Look at arctan1 x, 2 1 x - 22 x -2

First of all, the value range of tanx is to take the whole real number r, then its inverse function arctanx defines the domain is the whole real number r, then arctan1 x defines the domain, as long as the function is meaningful, that is, x≠0.

Its main basis:

The denominator of a fraction cannot be zero.

The number of open squares of the even square root is not less than zero.

The true number of the logarithmic function must be greater than zero.

The base of the exponential and logarithmic functions must be greater than zero and not equal to 1.

The domain of the inverse trigonometric function.

1. Arcsine function.

The inverse function of the sine function y=sin x on [- 2, 2] is called the arcsine function. Denoted as arcsinx, it represents an angle with a sinusoidal value of x, which is within the range of [- 2, 2]. Define the domain [-1,1] and the value range [-2, 2].

2. Inverse cosine function.

The inverse function of the cosine function y=cos x on [0, ] is called the inverse cosine function. Denoted as arccosx, it represents an angle with a cosine value of x, which ranges within the range of [0, ] . Define the domain [-1,1] and the value range [0, ].

Summary: y=arccosx is the inverse function of y=cosx(x [0, ], so the d domain of Arccosx is the domain of y=cosx(x [0, ].

The domain of definition refers to the range of values of the independent variable x, which is one of the three elements of the function (definition domain, value range, and corresponding law), and the corresponding object of the law.

There are three main types of questions for finding the function definition domain:

Abstract Functions Missing Numbers, General Functions, Function Application Problems.

Inverse function formula Hungry Brother.

1、cos(arcsinx)=√1-x^2）

2、arcsin(-x)=-arcsinx

3、arccos(-x)=πarccosx

4、arctan(-x)=-arctanx

5、arccot(-x)=πarccotx

6、arcsinx+arccosx=π/2=arctanx+arccotx

7、sin(arcsinx)=cos(arccosx)=tan(arctanx)=cot(arccotx)=x

Summary. Think of the function as an equation: y=f(x) solves the equation and finds the expression that x is identified by y, x=f (-1)(y) and swaps x,y to get the inverse function expression:

y=f (-1)(x) For example: find the inverse function of y=3x+5, the function is monotonic in (-, and the range is: (- So the domain of the inverse function is :

The range is: (-Solved by y=3x+5: x=1 3*y-5 3

Yes. Find a monotonic interval that is the domain of the inverse function.

Think of the function as an equation: y=f(x) Solve the equation and find the expression that x is identified by y, x=f (-1)(y) Swap x,y to get the inverse function expression: y=f (-1)(x) For example:

Find the inverse function of y=3x+5, the function is monotonic in (- , the range is: (- So the domain of the inverse function is: (megaroll- , the range is:

With guess prudence- , which is solved by y=3x+5: x=1 3*y-5 3

So the domain of the inverse function is: (-The range is: (-is solved by y=3x+5:

x=1 3*y-5 3 The inverse function is: y=1 3*x-5 3 x (-e.g. y=x 2, x=positive and negative root number y, then the inverse function of f(x) is the positive and negative root number x, pay attention to the definition domain and value range after the co-judgment is completed, the definition of the inverse function is the value range of the original function, and the value range of the inverse function is the definition domain of the original function. Debate.

This question has nothing to do with inverse functions. It's a composite function problem!

Since we finally let u=x+1 x

Thus f(u)=1 (u-2).

f(x)=1/(x²-2)

And the range of u=x+1 x is u-2 or u2 [because of the inequality of the mean].

y=[1+ (1-x)] 1- (1-x)] defines the domain: 1-x 0, x 1;1-√（1-x）≠0，x≠0；

Value range: set t= (1-x) 0,≠1

y=（1+t）/（1-t）

y-yt=1+t

y-1=t（y+1（

t=（y-1）/（y+1）≥0

y 1 or y -1

y=-1, the above denominator is meaningless, so y≠-1

The value range y 1, or y -1 is correct.

This question has nothing to do with inverse functions. It's a composite function problem!

Since we finally let u=x+1 x

Thus f(u)=1 (u-2).

f(x)=1/(x²-2)

And the range of u=x+1 x is u-2 or u2 [because of the inequality of the mean].