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Senior one mathematics
A domain is a value that can be obtained in parentheses of y=f(x).

If a represents the domain in the form of y=f(x), b represents the range.

I will explain why I mentioned the domain in the form of y=f(x) when I mentioned the domain earlier. )

The essence of function F is the mapping from one point in set A to another point in set B. You can also understand this mapping as a "function". When F acts on any point in A, it becomes any point in B. 。

For two different functions, f(x- 1) and f( 1/x+2), their values are different. Why? You should be able to understand them in essence, because their function values are all the same function f "acting" on different points.

x- 1

and

1/x+2

In fact, the function values obtained are of course different.

Moreover, their domains are different, such as f(x) and f(x/2), the latter domain is twice as large.

Why? I tell you this question in essence.

Now you have understood that the essence of function F is that it acts on any point in set A and becomes another point in set B. 。

So in this process, one thing remains the same:

That's a and b,

F can only act on the point in A forever, and the generated point is always the point in B.

What does this sentence mean?

Let me explain f(x) and f(x/2) to you for simplicity.

Assuming that the domain of f(x) is (-1, 1), let's find the domain of f(x/2).

The domain of f(x) is (-1, 1).

,

Explain that f can only act on points in the set (-1, 1) (actually it is a).

So I said before that when the form becomes f(x/2), the domain may change, but one thing will not change, that is, F can only act on the set A (- 1, 1), that is, x/2 (here you should be able to regard X/2 as the point where F acts).

,

This is the value range of x in f(x/2), which is the domain of f(x/2). The forms of f(x) and f(x/2) are different, but you can find that the values in brackets of f () are the same, both of which are (-1, 668).

.

I hope you can understand that this is because the set of f will not change with the change of form.

Now it is considered that the set in brackets of f () must be constant to solve the above problems.

The domain of f(x- 1) is [-2,3].

,

What does this sentence mean?

As long as the concept is clear, we should understand that the domain refers to the value range of X (the understanding of this definition is very important), that is to say, X can be taken in brackets [-2,3], so the range that can be taken in brackets of F () is [-3,2].

For f( 1/x+2), we follow the principle that the set in brackets of f () must be constant.

The range of 1/x+2 must also be [-3,2], so we can find the range of x based on this.

x∈(-∞,- 1/5)

Similarly, if you have a clear concept of domain, you should know that this is the domain of f( 1/x+2).

Summary:

1.

Function is the function of f on set A, so that these points correspond to (or become) another point in set B, respectively.

2.

For f(x- 1) and f( 1/x+2), the function values are different.

But the function of f is the same,

Why are the function values different? Because F acts on different points, x- 1 and 1/x+2, the points of "change" are naturally different.

In this process, the brackets of f () (that is, the range where f can be used) remain unchanged.

3.

We should correctly understand the true meaning of domain. Although the set in brackets of f () remains unchanged, it doesn't mean that.

f(x- 1)

and

f( 1/x+2)

The definition domain of is unchanged,

A function, no matter how it changes in parentheses, such as the more complex f (

x^3

+

x^2

+

3000 times

-

258/x

+ 1 19

),

Its domain is the value range of x,

We need to find the range of the set in brackets, that is, the range of x- 1 (set a), and define it according to other forms such as f(x- 1), and then make (

x^3

+

x^2

+

3000 times

-

258/x

+ 1 19

)

Still set a, so you can solve f (

x^3

+

x^2

+

3000 times

-

258/x

+ 1 19

The range of x).

4.

X in different functions f(x- 1) and f( 1/x+2) is just a symbol. You change the first function to f(t- 1).

,

Change the second one to f( 1/u+2),

This does not affect the solution process of function definition domain at all.

The second person's solution process is wrong here (but it seems to have been deleted by him. . . ), he is still very unclear about the whole mechanism and basic concepts of functions.

By the way, 1 this person's mistake. When seeking inequality, basic skills are still not enough. If you are not sure, you should take a special case from the result you seek to see if it meets the result.

correct

I'm glad to answer the landlord's question.

Please forgive me if there is any mistake.