Given that f (x) belongs to R for any x, f (- x) + F (x) = 0, f (x + 1) = f (x-1), then what is f (2011) equal to

Given that f (x) belongs to R for any x, f (- x) + F (x) = 0, f (x + 1) = f (x-1), then what is f (2011) equal to

F (x + 1) = f (x-1), let x = t + 1 get f (T) = f (T + 2) get f (x) take 2 as cycle, so f (2011) = f (1 + 2010) = f (1) and f (2011) = f (- 1 + 2012) = f (- 1), so 2F (2011) = f (1) + F (- 1) and because f (- x) + F (x) = 0, let x = 1 get f (1) + F (- 1) = 0, so 2F

If f (x) = (x-1) (X-2) (x-3)... (x-2011), f '(2011) =?
How to get f '(2011) = 2010

f(x)=g(x)(x-2011),g(x)=(x-1)..(x-2010)
f'(x)=g'(x)(x-2011)+g(x)
f'(2011)=g(2011)=2010*2009**..1=2010!

Given the function f (x) = | x + 1 | + | x + 2 | +... + | x + 2011 | + | - X-1 | + | - X-2 | +... + | - x-2011 |, why is it in [- 1,1]
Is it a constant function?

So | x + 1 | = | X - (- 1) | denotes the distance from X to - 1, and | X-1 | denotes the distance from X to 1. When x is in [- 1,1], | x + 1 | + | X-1 | is the distance from X to - 1 plus the distance from X to 1

According to the law, calculate 1 / 1 by 4 + 1 / 4 by 7 + 1 / 7 * 10 + +1 / 2008 * 2011
Rules: 1 / 1 * 4 = 1 / 3 (1-1 / 4), 1 / 4 * 7 = 1 / 3 (1 / 4-1 / 7), 1 / 7 * 10 = 1 / 3 (1 / 7-1 / 10) ,1/n（n+3）=1/3（1/n-1/n+3）

1 / 1 by 4 + 1 / 4 by 7 + 1 / 7 * 10 + +1/2008*2011
=1/3*(1-1/4+1/4-1/7+1/7-1/10+...+1/2008-1/2011)
=1/3*(1-1/2011)
=1/3*2010/2011
=670/2011