f(x)=f(x+1) for all real x, if f is a polynomial and f(5)=11,
then f(15/2) is ?
A. -11
B. 0
C. 11
D. 33/2
E. not uniquely determined
suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0
Posted: Tue Oct 14, 2008 1:45 pm Post subject:
Quote:
suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0
This is wrong. Fundamental theorem of algebra does not guarantee the existence of roots over field of rational numbers.
Consider for example p(x) = (x-5)^2+11.
Let degree n>0, then p(x) should go to +inf as x goes to +inf which cause contradiction. Therefore, p(x) must be a constant function.
vonLipwig wrote:Yes, it's that "exactly one smaller" which is important here. If you use that fact, then you don't need to consider the zeros of anything. You can just say that f(x+1) - f(x) is zero, so f(x) is degree 0, so constant.
Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”
Users browsing this forum: blahquaker, Bottomology and 4 guests