If the domain is symmetric about the origin, it is odd. If the domain is not symmetric about the origin, is it even?

If the domain is symmetric about the origin, it is odd. If the domain is not symmetric about the origin, is it even?

1. The domains of even and odd functions are symmetric with respect to the origin, and the functions that do not want to be symmetric with the origin are non odd and non even functions. The origin refers to the intersection of the x-axis and the y-axis, that is, the point where the number axis X = 0
2. The image of the even function is symmetric about the Y axis, and the image of the basis function is symmetric about the origin
No, and odd functions don't mean that the domain is symmetric about the origin
On the origin symmetry is odd function, on the Y axis symmetry is even function! A function is not symmetric about the origin, it may not be even, it may not be odd or even
No, even functions emphasize that the domain of definition is symmetric about the Y axis, and the Y coordinates are equal when the X coordinates are opposite. Specific questions can be asked, but they should be adopted.
No, it's even about X-axis symmetry
What is the condition that "f (x) is an even function" is that "the domain of definition of F (x) is symmetrical about the origin"
It's not necessary. Please tell me why
If f (x) is an even function, the answer must be the same when x is a positive or negative number. Therefore, the domain of F (x) must be symmetric about the origin, that is, the sufficient condition that "f (x) is an even function" is that "the domain of f (x) is symmetric about the origin"
The domain of F (x) is symmetric about the origin. For example, the domain of a function x belongs to R, which is symmetric about the origin. Can you say that it must be an even function? No! So the domain of F (x) is not symmetric about the origin, and f (x) is an even function
So "f (x) is an even function" is a necessary and sufficient condition for "the domain of F (x) is symmetric about the origin"
Senior high school mathematics problem: let the set of real numbers s be a set satisfying the following conditions: (1) 1 ∈ s; (2) if a ∈ s, then (1-A) / 1 proves that if a ∈ s
High school mathematics problem: let the real number set s be a set satisfying the following conditions: (1) 1 ∈ s; (2) if a ∈ s, then (1-A) / 1 prove that if a ∈ s, then 1 / 1-A ∈ s is 1 / 1-A ∈ S. similarly, according to the known conditions, take 1 / 1-A as a and substitute it into 1 / 1-A ∈ s, then we can get the conclusion: 1 / 1-1 / 1-A ∈ s
It is proved that if a ∈ s, then 1 / 1-A ∈ s
According to the conditions
② If a ∈ s, then 1 / 1-A ∈ s
Let t = / 1-A ∈ s
If t ∈ s, then 1 / 1-T ∈ s
Isn't it 1 / 1-A ∈ s that leads to 1 / 1-1 / 1-A ∈ s?
If the function f (x) = x2-4x + 3, the set M = {(x, y) | f (x) + F (y) ≤ 0}, and the set n = {(x, y) | f (x) - f (y) ≥ 0}, then the area of the region represented by the set M ∩ n in the plane rectangular coordinate system is______ .
Because f (x) = x2-4x + 3, f (y) = y2-4y + 3, then f (x) + F (y) = (X-2) 2 + (Y-2) 2-2, f (x) - f (y) = x2-y2-4 (X-Y) = (X-Y) (x + y-4).. P = {(x, y) | (X-2) 2 + (Y-2) 2 ≤ 2}, q = {(x, y) | (X-Y) (x + y-4) ≥ 0} π．
Given the real number A & sup2; - A + 1,3, a, - 1, the set composed of objects is m, and there are only three elements in M, find the set composed of such different real numbers a
Because there are only three elements in the M set,
So, a2-a + 1 = 3, a2-a + 1 = a, a2-a + 1 = - 1, a = 3, a = - 1
The solution is: a set is {- 1,1,2,3}
If the function f (x) = cos (3x + φ) is odd, then the value of φ is
π/2+2kπ
Odd function f (- x) = - f (x)
cos(-3+φ)=-cos(3x+φ)
Let x = 0
cos(φ)=-cos(φ)
0=-0
φ=π/2+2kπ
One thing about odd functions is that f (0) = 0. You can get it by bringing x = 0 in
Let a real number A & sup2; - A + 1,3, a, - 1 be a set M of elements, and M contains only three elements
Then how many different numbers of a are there
Isn't it three, - one, one, two? These four
- 1 why not
There are three: a = 1: when a & # 178; - A + 1 = a, the set is {3,1, - 1}, which meets the requirements; 3: when a = 3, so a & # 178; - A + 1 = 7, the set is {3,7, - 1}, which meets the requirements; 2: when a & # 178; - A + 1 = 3, a = 2 or - 1, when a = 2, the set is {3,2, - 1}, which meets the requirements; when a = - 1, the set is {3, - 1}
That's all four
Three, three, two, one
Function f (x) = cos (3x + φ - π / 6) (0
F（0）=0
cos(φ-π/6)=0
φ-π/6=π/2+kπ
Another 0
1. If the set M is composed of real numbers A & sup2; - A + 1,3, a, - 1, and m only contains three elements, how many values of a are there in different real numbers a
When a = - 1, a & # 178; - A + 1 = 3, then there are two elements in M, which are rounded off,
When a = 3, a & # 178; - A + 1 = 9-3 + 1 = 7, there are three elements in M,
When a & # 178; - A + 1 = a, a = 1, there are three elements in M,
When a & # 178; - A + 1 = - 1, a = 0 or a = 1. If a = 0, then M has four elements. If a = 1, it has been discussed,
When a & # 178; - A + 1 = 3, a = 2 or a = - 1, if a = - 1, which has been discussed, is not tenable, and is discarded. If a = 2, then there are three elements in M, tenable,
To sum up, there are three values of a that can make the title hold, namely: 1, 2, 3,
If the function f (x) = cos (3x, φ) is odd, then the minimum positive value of φ is
The function f (x) = cos (3x + φ) is odd
So f (0) = cos φ = 0
So φ = k π + π / 2, K ∈ Z
Then the minimum positive value of φ is π / 2
If you don't understand, I wish you a happy study!
Explanation
F (x) = cos (3x + φ) is an odd function,
Then f (x) can be converted to f (x) = ± sin3x
So φ = k π + π / 2, K belongs to Z
So when k = 0, there is a minimum positive value of π / 2