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.