##### Question 1.8
lydiark25
I'm confused on the answer for this question's answer. How is it that f(1)=0, f(2m)=m, and f(2m+1)= -m and how does that answer the problem? Thank you! -Lydia Kirillova
weisbart: Jan. 15, 2015, 5:33 p.m.
Try plugging in numbers. What is $f(1)$, $f(2)$, $f(3)$, $f(4)$, $f(5)$, $f(6)$, and $f(7)$? Anyone want to try it?
martha2A: Jan. 15, 2015, 11:13 p.m.
f(1)=0 f(2)=1 f(3)=-1 f(4)=2 f(5)=-2 f(6)=3 f(7)=-3 These seem to follow a pattern of the even x-values having positive y-values and the odd x-values having negative y-values
martha2A: Jan. 15, 2015, 11:26 p.m.
I think the answer would be f(2m)=m for even x-values and f(2m-1)=-m+1 for odd x-values. f(1)=0 because, using the function for the odd x-values, we can get one inside the parentheses by letting m=1 which then gives us f(1)=0.
weisbart: Jan. 18, 2015, 10:58 p.m.
You're right Martha! That works fine and I like your answer better than the one I gave. The one I gave will also work, but it requires an additional condition. That's why I have three conditions--$f(2m+1)$ only gives the function for the odd numbers starting with 3, I need an additional condition for $f(1)$. In the end, it doesn't make much difference because the function is the same, but when you write it the way you did it is a little simpler.