Did you try a graph?

Results 1 to 4 of 4

- October 20th 2008, 04:45 AM #1

- Joined
- Sep 2008
- Posts
- 222

## [SOLVED] x-[x]?

Hey,

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

- October 20th 2008, 08:59 AM #2

- October 25th 2008, 05:45 PM #3

- Joined
- Sep 2008
- Posts
- 222

- October 26th 2008, 04:25 AM #4

- Joined
- Apr 2005
- Posts
- 17,190
- Thanks
- 2097

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.