I know that the domain of definition refers to the scope of X. what does "defined in" mean also refers to the overall scope of X or F () in brackets. Beginners don't know much about it,

I know that the domain of definition refers to the scope of X. what does "defined in" mean also refers to the overall scope of X or F () in brackets. Beginners don't know much about it,

You need to understand the definition of a function. The X in the function domain must have a value pre support correspondence in the value domain y, and it must be "one to many" or "one to one". Your domain is right, but "defined in" does not refer to the definition of X in X or F (), which only specifies that the domain of X is in the range of "defined in", It means that the domain of definition is in R. bin is not the domain of definition, but R
It means the value range of X.
The two concepts are the same
Domain refers to the value range of the independent variable. It depends on what the independent variable is
When it comes to functions, there must be domain and range
Both of them are essential elements of function
Definition in... Is actually to describe what the domain of a function is
For example. For example, if the definition field of function f (x) = x is r, we will encounter the function f (x) = x defined on [- 1,1], then we can say that the definition field is the value range of x where the function can exist. For example, the definition field of F (x) = 1 / X is {x | x ≠ 0, X ∈ r}
For example, the function f (x) = x defined on [- 1,1] can be understood in this way: [- 1,1] is not a domain, but it is a part of the domain. And we are going to solve the problem in X ∈ [- 1,1]. ... unfold
For example. For example, if the definition field of function f (x) = x is r, we will encounter the function f (x) = x defined on [- 1,1], then we can say that the definition field is the value range of x where the function can exist. For example, the definition field of F (x) = 1 / X is {x | x ≠ 0, X ∈ r}
For example, the function f (x) = x defined on [- 1,1] can be understood in this way: [- 1,1] is not a domain, but it is a part of the domain. And we are going to solve the problem in X ∈ [- 1,1]. Put it away
The sixth power of a = the second power of () the second power of (A's fifth power) multiplied by the second power of () the eighth power of a multiplied by the sixth power of a
The 6th power of a = the 2nd power of (A & # 179;); the 2nd power of (A's 5th power) multiplied by the 2nd power of (A & # 178;) = the 8th power of a multiplied by the 6th power of A
(2010, Anhui) if f (x) is an odd function with period 5 on R, and f (1) = 1, f (2) = 2, then f (3) - f (4) = ()
A. 1B. 2C. -2D. -1
∵ if f (x) is an odd function with period 5 on R ∵ f (- x) = - f (x), f (x + 5) = f (x), ∵ f (3) = f (- 2) = - f (2) = - 2, f (4) = f (- 1) = - f (1) = - 1, ∵ f (3) - f (4) = - 2 - (- 1) = - 1
The 8th power of T - the 2nd power of T multiplied by the 4th power of T=
The eighth power of T - the second power of T multiplied by the fourth power of T
=t^8-t^6
=t^6(t^2-1)
Odd function and f (- x) = f (X-2), what is the period
Four
Fill in the blanks with factorization factor: 9th power of a + 7th power of a - 6th power of 2A - 5th power of 3A = () (4th power of a + 2nd power of a - 2a-3)
The 9th power of a + the 7th power of a - the 6th power of 2A - the 5th power of 3A = (the 5th power of a) (the 4th power of a + the 2nd power of a - 2a-3)
(2010, Anhui) if f (x) is an odd function with period 5 on R, and f (1) = 1, f (2) = 2, then f (3) - f (4) = ()
A. 1B. 2C. -2D. -1
∵ if f (x) is an odd function with period 5 on R ∵ f (- x) = - f (x), f (x + 5) = f (x), ∵ f (3) = f (- 2) = - f (2) = - 2, f (4) = f (- 1) = - f (1) = - 1, ∵ f (3) - f (4) = - 2 - (- 1) = - 1
(2a-b + C) ^ 2 - (3a + 2b-c) ^ 2 factorization
F (x) is an odd function defined on R with a period of 3, and f (1) = 7
F (x) is an odd function defined on R with period 3,
f(-x)=-f(x)
f(x)=f(x+3)
F(1)=7=f(-2)=f(-5)=-f(5)
f(5)=-7
Factorization (2a + b) ^ 2 - (3a-2b) ^ 2