Question

range of (2x^2)/(x^2-16)

Answer 1

(2x^2)/(x^2-16) = (2x^2)/[(x+4)(x-4)]

since the denominator cannot be 0, that implies that:
x =/= 4, -4

so the domain is {x|x=/=4,-4} in other words, x can be anything except 4 and -4.

for the range:
the range is all of the real numbers, assuming x is real as well.
since if x get be infinitely positive -> range can be infinitely positive
likewise if x get infinitely negative -> range can be infinitely negative
finally x=0 -> range contains 0

Answer 2

oops lol, for the range part of my post i made a mistake...let me correct it...

Answer 3

range can be infinitely negative not because x can get infinitely negative, but infinitely close to 0, can you see why?

Answer 4

mk. mind giving me the answer in interval notation as well? the program im using only accepts interval notation

Answer 5

@findme, what would the answer be in interval notation?

Answer 6

(-infinity, +infinity)

Answer 7

tried that already, its not the right answer

http://gyazo.com/4aed0195929f41c86c9bbd8c1c859129

Answer 8

oops, wrong equation lol....
anyways
the equation is (2x^2)/(x^2-16)
so the range is:
(-inf,0]U(2, inf)

Answer 9

correct, thank you.

Answer 10

np, sorry about the mistakes, keep on thinking about previous questions.....