Define the odd function f (x) of domain R, when x ∈ (－∞, 0), f (x) + XF '(x)

Define the odd function f (x) of domain R, when x ∈ (－∞, 0), f (x) + XF '(x)

Answer:
F (x) is an odd function, f (- x) = - f (x)
G (x) = XF (x), the definition field is the same as f (x), both of which are real number ranges R, symmetrical about the origin
g(-x)=(-x)f(-x)=-x*[-f(x)]=xf(x)=g(x)
So: G (x) is an even function
g'(x)=f(x)+xf'(x)

For the odd function f (x) whose domain is r, f (x) · f (- x) = 0 is always true?

Odd function f (x) with domain R
f(-x)=-f(x)
f(x)·f(-x)=0=-[f(x)]^2=0
f(x)=0
I don't know if you want this. If it's an implicit function, it's correct. If it's an explicit function, it's incorrect

For any odd function f (x) whose domain is r, the following relationship is always true? A. F (x) - f (- x) is greater than or equal to 0 B.F (x) - f (- x) light rain or equal to 0 C.F (x) times f (- x) light rain or equal to 0 D.F (x) times f (- x) is greater than or equal to 0,

C
From the odd function f (x) = - f (x)
Term a & term B, f (x) - f (- x) = 2F (x) positive and negative indeterminate
Term D, f (x) multiplied by F (- x) = - [f (x)] square

It is known that the function f (x) is an odd function with the definition field R, and is constant for any x ∈ R, f (1-x) = f (1 + x). Find the value of F (0) (1) Find the value of F (0), and prove that f (x) is a periodic function with a period of 4; (2) If f (x) = x when x ∈ (0,1], find the analytical formula of function f (x) when x ∈ [- 1,1]

The function f (x) is an odd function whose domain is r, so f (0) = 0f (x) = - f (- x) f (1-x) = f (1 + x) makes t = 1-x x = 1-T, so f (T) = f (2-T) = - F (- t) makes - t = a, so f (a) = - f (a + 2) f (a + 2) = - f (a + 4), so f (a) = f (a + 4) f (x) is a periodic function with a period of 4; Let x ∈ [- 1,0) - x ∈ (0,1] f (x)

It is known that FX is an even function defined on R. for any x € R, f (x + 6) = f (x) + 2F (3) and f2013 It is known that FX is an even function defined on R. for any x € R, f (x + 6) = f (x) + 2F (3), then f (2013) is equal to

Let x = - 3 be substituted into f (x + 6) = f (x) + 2F (3)
f（3）=f（-3）+2f（3）
And f (x) is an even function, so f (3) = f (- 3)
So we get f (3) = 3f (3)
f（3）=0
So f (x + 6) = f (x)
So f (2013) = f (335 * 6 + 3) = f (3) = 0

It is known that f (x) is an even function defined on R. for any x ∈ R, f (x + 6) = f (x) + 2F (3), f (- 1) = 2, then f (2011) = () A. 1 B. 2 C. 3 D. 4

F (x + 6) = f (x) + 2F (3), and f (x) is an even function defined on R
Let x = - 3 give f (3) = f (- 3) + 2F (3) and f (- 3) = f (3)
∴f（-3）=f（3）=0
‡ f (x + 6) = f (x), that is, the function is a function with a period of 6
∵f（-1）=2
∴f（2011）=f（1）=f（-1）=2
So choose B