f(x+1) ≥ f(x) + 1
f(x+5) ≤ f(x) + 5
g(x) = f(x) + 1 - x
find g(2008)
1We easily get f(x+5)=f(x)+5 by considering double inequalities about f(x+5).
that implies f(x+1)=f(x)+1.
let f(1)=c.
implying, f(x+1)=c+x for all x ε Z.
then, g(2008)=1-2008+2007+c=c.
also g(x) is a constant function.
if any val. of g is given(at any integral x) then we would find value for c.
1from f(x+1)=f(x)+1
we can also write
f(x) = [x] (im just saying from that piece of info forget the rest conditions!!)
SO THE QUESTION IS INCOMPLETE IN MANY WAYS!!
62yes rohan precisely.. i deleted my post in between bcos i din want to interfere in ur healthy discussions...
but gr8 u got the ponit..
actually u need to know f(x) or g(x) or either for an interval of lenght 1..
then ;u can uniquely determine the values!
62but bcos we need to find f(2008) we need to know f at 0 or some integer..
i think that will suffice!
1f(0)=0 was given.. sorry i forgot that.
but i understood the solution..
thanks :)