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3 answers

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

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