Firts, lets consider non negative values of x. Then,f is defined for every x >1 that's not an integer. For x in [0, 1], [x] = 0 and f is undefined.
So, for each positive integer n, we have
f(x) = x/n for x in (n, n+1), that is,
f(x) = x for 1 < x<2
f(x) = x/2 for 2 < x <3
f(x) = x/3 for 3 < x < 4, and so on.
In the n-th iterval, f is increasing and f(x) is in (1, (n+1)/n) = (1, 1 + 1/n). So, the possible values for f (range of f) is the set
(1,2) U (1,1/2) U (1, 2/3) U (1 + 3/4).....U (1 , 1 + 1/n)
Since all these sets are contained in (1, 2), their union is (1,2).
Since f is an even function, it's the same on the negative axis. Hence, the set of all possible values for f is (1,2).
And the domain of f is R - (-1,1) - Z , Z the set of integers
2007-12-20 06:39:18
·
answer #1
·
answered by Steiner 7
·
0⤊
0⤋
The possible values are
1. 0 ( assuming 0/0 = 0 ) ,
2. infinity ,
3. all values between 1 and 2 excluding the value 2 , and
4. all values between -1 and -2 excluding the value -2.
2007-12-20 11:31:17
·
answer #2
·
answered by NARAYAN RAO 5
·
0⤊
2⤋
The domain (you mean?) is (-2, -1 ] U [1, 2)
2007-12-20 11:27:03
·
answer #3
·
answered by lienad14 6
·
0⤊
1⤋