h(x)=x^(1/2) + (x+1)/(x-1) + 3x
2007-09-10 07:07:30 · 5 answers · asked by jducksgolf 2 in Science & Mathematics ➔ Mathematics
x >= 0 and x different of 1
2007-09-10 07:11:34 · answer #1 · answered by antone_fo 4 · 0⤊ 0⤋
0 to 1 and from 1 to positive infinity
2007-09-10 07:13:14 · answer #2 · answered by gebobs 6 · 0⤊ 0⤋
There are 2 parts to this
first x^.5 is imaginary below Zero
second 1/(x-1) is assymptotic at 1
so the answer would be:
[0,1) and (1,infinity]
2007-09-10 07:13:48 · answer #3 · answered by Sugar Shane 3 · 0⤊ 0⤋
This function in discontinuous at x = 1 where the denominator of the second term would be zero. (The function does not exist for values of x less than zero if this is a function of real variables.)
2007-09-10 07:16:42 · answer #4 · answered by sirena 1 · 0⤊ 0⤋
Well, it's imaginary for x < 0
It's undefined/infinite for x = 1
Other than that, it's all right.
2007-09-10 07:13:44 · answer #5 · answered by PMP 5 · 0⤊ 0⤋