Question

Find the domain of (sqrt(3x+12))/(x^2-36)

Answer 1

[-4,6) U (6,∞)

Answer 2

How did you get 4?

Answer 3

This is what I get x^2-36
x^2-36=/=0
x^2 =/= 36
x =/= +- 6
(-inifinity, -6) (6,infinity)

Answer 4

\[\left( -\infty , 6 \right) \upsilon \left( 6 , \right)\]

Answer 5

(x-6)(x+6)=0
(3(x+4))<0

(x-6)(x+6)=0

x=6,-6

Solve for x

(3(x+4))<0

x<-4

x≠6 x≠-6 x≥-4

[-4,6) U (6,∞)

Answer 6

Why does the numerator have restrictions?

Answer 7

i think yesterday is right

Answer 8

you do not want to take the square root of a negative number

Answer 9

oh right lol

Answer 10

Ok so 3x+12 > 0
3x > -12
x > -4
Why is it not
(-4,-6)(6,infinity)?

Answer 11

that looks ok. means the same thing.

Answer 12

Oops i mean >=
[-4,-6) (6,inifinity)
How did you get only positive 6?

Answer 13

How did you get (-6,6)?

Answer 14

check that
(-4,6)(6,inf)

Answer 15

Wow I have no idea what you're saying :(
Why don't you include -6 in (-4,6)(6,inf)

Answer 16

because -6 is less than -4, so it is already excluded

Answer 17

Oooh that makes sense, but why (-6,6)?

Answer 18

-6 and 6 are excluded because they would cause division by zero

Answer 19

Alright nvm thanks! I figured it out. It includes -4 right?

Answer 20

and to be totally correct you should write
[-4,6)(6,inf)
because -4 is allowed

Answer 21

which is mitch's very first post up top.

Answer 22

Yeah, but he didn't explain it like you did :) Sorry Mitch I can only give one medal

Answer 23

np