Suppose that f(x+1)=f(x) for all real x. If f is a polynomial and f(5) = 11, the f(15/2) is:
a)11
b) 0
c)11
d) 33/2
e)not uniquely determined by the information given
I thought answer e) but actually it's c), I don't figure out why?
GRE 8767 #20

 Posts: 4
 Joined: Sun Oct 19, 2014 12:09 am
Re: GRE 8767 #20
The key is to realize that f is a polynomial and use the fact that f(5) = 11 with f(x) = f(x+1) for all real x to deduce it must be constant. Hence, you can obtain f(7.5).