Did you try a graph?
This is on the Barron's SAT Math II Book page 22 #50
If [x] is defined to represent the greatest integer less than or equal to x, and f(x)= l x-[x]-1/2 l what is the period of f(x)?
how do you do this problem? Please list the steps. Thanks in advance
f(x)= l x-[x]-1/2 l so f(0)= |-1/2|= 1/2. If f is periodic then there must be other values of x such that f(x)= 1/2. The period is the shortest distance from x= 0 to another such value of 1/2. Solve |x- [x]- 1/2|= 1/2.
1) Suppose x- [x]-1/2> 0. Then x-[x]- 1/2= 1/2 so x- [x]= 1. Since [x] is the largest integer less than or equal to x, x- [x] must be less than 1. That equation is never true.
2) Suppose x- [x]- 1/2< 0. Then x- [x]-1/2= -1/2 or x- [x]= 0. That is true if and only if x= [x] or x is itself an integer.
That is, f(x)= 1/2 for x any positive integer. It is periodic with period 1.