Find range of values

arze · #1 $Old$ November 1st, 2009, 10:21 PM

find the range of values for k for which the function
$\frac{x^2-1}{(x-2)(x+k)}$
where x is real, takes all real values.
I don't know how to start. Any pointers?
Thanks

statmajor · #2 $Old$ November 4th, 2009, 01:30 PM

The question asks you for what values of X, would F(x) exist. Or what values of X would F(x) not exist.

Hint: C/0 doesn't exist (where C is an arbitrary constant).

arze · #3 $Old$ November 4th, 2009, 05:55 PM

so x cannot be 2 or -k. i dont know what else the answer is $|k|\leq 1$

statmajor · #4 $Old$ November 4th, 2009, 06:10 PM

Sorry, I misread the question. I thought you were supposed to find values for x instead of k.

So same as idea as before. We know that k cannot be -x and we also know that x can't be 2. Therefore, k cannot be -2 (I think).

arze · #5 $Old$ November 4th, 2009, 08:29 PM

ok got that, but how to continue confuses me

statmajor · #6 $Old$ November 4th, 2009, 09:00 PM

What part are you confused about? You know that k can't be -x, and when x = 2, k can't be -2.

arze · #7 $Old$ November 4th, 2009, 09:01 PM

yes, i meant how to find the values of k other than that it isn't -2.

statmajor · #8 $Old$ November 4th, 2009, 09:19 PM

The domain for is k $(- \infty, -x)\cup(-x, \infty)$
The domain of x is $(- \infty, -2)\cup(-2, \infty)$