The function f (x) is an odd function defined on R, and when x ∈ (0, + ∞), f (x) = 2x, then f (log21 3)=______．

The function f (x) is an odd function defined on R, and when x ∈ (0, + ∞), f (x) = 2x, then f (log21 3)=______．

∵ function f (x) is an odd function defined on R
∴f(log21
3)=-f（log23）
∵ when x ∈ (0, + ∞), f (x) = 2x,
∴f（log23）=2log23=3
∴f(log21
3)=-3
So the answer is: - 3

The function f (x) defined on R satisfies that f (x) = log2 (1-x) x ≤ 0; F (x) = f (x-1) - f (X-2) x > 0, find f (2011) Find f (2011) = - f (2008). What's the next calculation?

f(x)=f(x-1)-f(x-2) f(x-1)=f(x-2)-f(x-3) f(x)=-f(x-3) f(x-3)=f(x-4)-f(x-5) f(x-4)=f(x-5)-f(x-6) f(x)=f(x-6) f(2011)=f(1+335*6)=f(1)=f(0)-f(-1)=log2(1-0)-log2(1-(-1))=0-1=-1

If the odd function f (x) defined on R satisfies f (x-3) = f (x + 2) and f (1) = 2, then f (2011) - f (2010) = __

∵ f (x) satisfies that f (x-3) = f (x + 2), ∵ the period of the function is 5,
∵ odd function f (x), f (1) = 2, ∵ f (- 1) = - 2, f (0) = 0 on R
∴f（2011）-f（2010）=f（1）-f（0）=2-0=2
So the answer is 2

It is known that the function f (x) defined on R is an odd function, and the period of function f (2x + 1) is 5. If f (1) = 5, the value of F (2009) + F (2010) is () A. 5 B. 1 C. 0 D. -5

∵ the period of function f (2x + 1) is 5 ∵ 2 (x + 5) + 1] = f (2x + 1), that is, f (2x + 11) = f (2x + 1), that is, f (y + 10) = f (y), so the period of function f (x) is 10 ∵ f (2009) = f (- 1), f (2010) = f (0) ∵ function f (x) is an odd function defined on R ∵ f (0) = 0, f (- 1) = - f

The odd function f (x) of the definition field on R satisfies that f (x-3) = f (x + 2) and f (1) = 2 find f (2011) - f (2010) The odd function f (x) of the definition field on R satisfies that f (x-3) = f (x + 2) and f (1) = 2 find f (2011) - f (2010)

Odd function f (- x) = - f (x),
X = 0, then f (- 0) = - f (0), f (0) = 0
F (x-3) = f (x + 2), let x-3 = a,
F (a) = f (a + 5), the period of the function is 5
f（2011）-f(2010)= f（2011-5*402）-f(2010-5*402)
=f(1)-f(0)=2-0=0.

The definition field of function f (x) is R. if both f (x + 1) and f (x-1) are odd functions, and when x belongs to [0,1], f (x) = log2 (2-x), then f (2010) + F (2011) What is the value of

F (x + 1) = - f (- x + 1), f (x-1) = - f (- x-1), then
F [(x + 2) - 1] = f [- (x + 2) - 1] = - f (- x-3) = - f (- x + 1), that is, f (- x-3) = f (- x + 1),
So the period of F (x) is 4,
f(2010)+f(2011)=f（4x502+2）+f（4x503-1）
=f（2）+f（-1）,
f（2）=f（1+1）=f（1-1）=f（0）=log2 2=1,
F (- 1) = f (0-1) = - f (- 0-1) = - f (- 1), then f (- 1) = 0,
So f (2010) + F (2011) = f (- 1) + F (2) = 0 + 1 = 1