Calculation and meaning in definition field of high school function It is known that the definition of function y = f (x) is [0,1]. ① find the domain of y = f (X & sup2;) and ② find the domain of function y = f (x / x + 1) When we find ①, we find that the function y = f (X & sup2;) satisfies 0 ≤ X & sup2; ≤ 1, but then how can we get - 1 ≤ x ≤ 1? When we find the function y = f (x / x + 1) satisfies 0 ≤ X / x + 1 ≤ 1, but how can we get x ≥ 0? Sorry, I just began to learn not to understand, hope you teach me

Calculation and meaning in definition field of high school function It is known that the definition of function y = f (x) is [0,1]. ① find the domain of y = f (X & sup2;) and ② find the domain of function y = f (x / x + 1) When we find ①, we find that the function y = f (X & sup2;) satisfies 0 ≤ X & sup2; ≤ 1, but then how can we get - 1 ≤ x ≤ 1? When we find the function y = f (x / x + 1) satisfies 0 ≤ X / x + 1 ≤ 1, but how can we get x ≥ 0? Sorry, I just began to learn not to understand, hope you teach me

① 0 ≤ X & sup2; ≤ 1
The solution set of 0 ≤ X & sup2; is r
The solution set of X & sup2; ≤ 1 is - 1 ≤ x ≤ 1
0 ≤ X & sup2; ≤ 1 is equivalent to the system of inequalities X & sup2; ≤ 1 and X & sup2; ≥ 0
The solution set of the inequality system X & sup2; ≤ 1 and X & sup2; ≥ 0 is the intersection of R and - 1 ≤ x ≤ 1,
Namely:
-1≤x≤1
Therefore, the solution set of inequality 0 ≤ X & sup2; ≤ 1 is - 1 ≤ x ≤ 1
② The inequality 0 ≤ X / x + 1 ≤ 1 is equivalent to the system of Fractional Inequalities X / x + 1 ≤ 1 and X / x + 1 ≥ 0
The solution set of inequality X / x + 1 ≥ 0 is x ≥ 0 or X ≤ - 1
Inequality X / x + 1 ≤ 1

Is it the same concept as a key university?

Universities are divided into three categories: undergraduate, junior college and vocational college
What you ask is undergraduate course. Undergraduate course is divided into one, two and three. One is divided into ordinary one, 211 university and 985 University. For key university, it is a relatively vague concept. Sometimes it refers to all one university. Sometimes it refers to a better one. There are many differences between the two. 211 university can be said to be a better one
To tell you more, 985 university is a higher level than 211 university, that is 985 > 211 > ordinary undergraduate (you say the key university). Then 985 university is 211 university, 211 university may not be 985 University. There are many 985 University and 211 university, that is the real key university~

Who knows the two definitions of chemical reaction rate? The definition of high school is different from that of University. The definition of high school is defined by IUPAC!
The two results are very different!

(1) Definition of chemical reaction rate --- the speed of chemical reaction!
(2) Average rate of chemical reaction --- change of concentration in unit time!
(3) Instant rate of chemical reaction -- the limit of infinitesimal T, the average rate of chemical reaction!

A question about the concept of function in Senior High School
If the domain of function y = f (x) is [- 2,4], then the domain of function g (x) = f (x) + F (- x) is?
I want a detailed process, thank you!

The definition field of function y = f (x) is [- 2,4], so the definition field of F (- x) is - 2

Conceptual problems of function in Senior High School
Power function, exponential function, logarithmic function. As far as the functions learned in senior high school are concerned, can they all perform four operations?
For example, can two power functions be subtracted in addition to addition and multiplication?
How about exponential function and logarithmic function!

All are OK ~ but some have no formula

High school -- the concept and nature of function
If f (x) = x & sup2; - 2 (1-A) x + 2 is a decreasing function in (- ∞, 4], then the set of values of real number a is?

Function symmetry axis 1-A > = 4

On the definition domain of high school function
The domain of ∵ f (x) is [2,4], that is, 1 ≤ x ≤ 2, and the domain of ∵ f (2x) is [1,2]
I can do this problem
Now the main question is, are f (x) and f (2x) the same function
If not, there is no reason for 2 ≤ 2x ≤ 4
If it is... The domain of F (2x) is different from that of F (x), so it is not the same function
It's strange. What's wrong with me=
Why is the de value range of X equal to the value range of 2x
I just want to know if these two functions are the same function.. = 1=
The range of independent variable of F (x) is 2 ≤ x ≤ 4
Therefore, when the independent variable x is 2x, the condition must be in the range of [2,4]
So the domain of F (2x) is [1,2]
The problem is that the domain of F (2x) is [1,2], so the range of X in F (2x) is 1 ≤ x ≤ 2
But it's different from the domain in the original function f (x)
I know that the domain corresponding to function equality is the same, but the domain is different here?
So I'm wondering
If I can understand

Y = f (x) is the original function
Y = f (2x) is a composite function and u = 2x is an intermediate function
When we define y = f (U), we mean that for the function rule F, the independent variable is u
However, when u is another function, the function rule will change, not the original F
For example, y = f (x) = 2 ^ x is an exponential function and an elementary function
However, y = 2 ^ (x + 2) is not an exponential function, just an exponential function, it is a composite function
It can also be understood that y = f (x + 2), and the intermediate function is u = x + 2
Obviously, y = 2 ^ X and y = 2 ^ (x + 2) are not the same as functions, because the rules of functions are different
Can you accept this explanation?
In addition, the three elements of a function are: domain of definition, range of values, and rule of function
Therefore, the necessary and sufficient condition for the two functions to be the same is that the domain of definition, the range of value and the rules of function are all the same
About your question: why is the value range of X equal to that of 2x
Function expression y = f (x), the independent variable is x, the selection of the independent variable letter has nothing to do with the function relationship, so even if
Y = f (T), which is the same range as X
However, for the function rule F of y = f (x), y = f (U) and so on, the independent variable is something in brackets, even if it is very complex. For example, in F (AX + b), for F, its independent variable is the whole block (AX + B) in brackets
So the range of F (x) and f (2x), X and 2x are the same
This is the difference between the original function and the composite function

The problem of definition field of function in Senior High School
Is the domain of function f (x + 1) [0,1] the range of X or the range of X + 1
I think the domain refers to the range of independent variables, that is the range of X + 1, because if t = x + 1, then the domain of F (x + 1) = f (T) is [0,1], that is the range of T

The domain of function f (x + 1) is [0,1], which refers to the range of X
For example: what is the independent variable of F (x + 1)? Is its corresponding rule or F? The independent variable of F (x + 1) is x, and its corresponding rule is not f. we may as well make the following assumption: if f (x) = x & # 178; + 1, then f (x + 1) = (x + 1) &# 178; + 1, f (x + 1) and (x + 1) &# 178; + 1 are equal, that is, the independent variable of (x + 1) &# 178; + 1 is the independent variable of F (x + 1). (x + 1) &# 178; +The corresponding rule of 1 is to first add 1 to the independent variable, then square it, and then add 1
For example: is the domain of y = f (x) 1 ≤ x ≤ 2 the same as that of y = f (x + 1)? Is the range of values the same? If we know that the domain of F (x) is x ∈ [1,2], what is the domain of F (x + 1)? Because the domain of F (x) is x ∈ [1,2], that is to say, every value f (x) in 1 ≤ x ≤ 2 has a function value, and any value f (x) beyond this range has no function value. For example, 3 has no function value, Therefore, when the value of X + 1 exceeds the range of [1,2], f (x + 1) has no function value, so the domain of F (x + 1) is the solution set of the inequality 1 ≤ x + 1 ≤ 2, that is to say, the domain of X + 1 in F (x + 1) is the domain of F (x), and since 1 ≤ x + 1 ≤ 2, the domain of F (x + 1) is naturally the same as that of F (x) (1 ≤ x ≤ 2)

Given that the domain of F (x) is (- 1,2), find the domain of y = f (2x + 3)
Would you like to take it directly
Or - 1

Yes

High school questions about the domain of function definition
Let the domain of function f (x) be [0,2], then the domain of function f (x-1) is ()
A.[0,2] B.[-1,1] C.[1,3] D.[-1,0]

Let X-1 in function f (x-1) be t, then the domain of definition of function f (x) is [0,2], with 0=