I know these are really simple, but for some reason I do not believe I am looking at these correctly...are these answers correct?
1) f(n) = +-n
No.
2) f(n) = SQRT(n^2 + 1)
Yes.
3) f(n) = 1/(n^2-4)
No.
2007-02-17 04:31:18 · 2 answers · asked by jaden404 4 in Science & Mathematics ➔ Mathematics
Z = all integers
R = all real numbers
Determine whether f is a function from Z to R if:
1) f(n) = +-n
2) f(n) = SQRT(n^2 + 1)
3) f(n) = 1/(n^2-4)
2007-02-17 05:07:59 · update #1
I don't totally agree with the first two replys.
Context and notation suggest that you mean:
Z is the integers, R is the reals.
1)
2) the sqrt threw me off at first. But I graphed it . That helped me see that it is a function.
3) The only problem is singularities at n = -2 and +2. It's still a function although it's not defined at those two values.
2007-02-17 04:59:15 · answer #1 · answered by modulo_function 7 · 0⤊ 0⤋
Nos. 1 and 2 are correct. Why?
No. 3 is wrong. For, suppose that two values of f(n) give the same values of n. Show that these two values of f(n) are the same.
I expect that you answer these questions and not take my word for it.
Edit: Scratch that, no. 3 is also right; why?
2007-02-17 12:40:18 · answer #2 · answered by Anonymous · 0⤊ 0⤋