Question 1088423
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Given Function:
<img src = "https://image.prntscr.com/image/mvFZwy4kTcOyPT_lzHnZXA.png">
Note: If the image doesn't show up, then it's the function g(x) = 1/(x(x+2))


The domain is the set of numbers x where x can be any real number but it cannot be equal to zero, and it also can't be equal to -2
Why can't x equal these values? Because either value makes the denominator x(x+2) equal to zero.


We can see this by solving x(x+2) = 0 for x
x(x+2) = 0
x(x+2) = 0
x=0 or x+2 = 0
x = 0 or x = -2


And we can check each value
Plug in x = 0
x(x+2) = 0
0(0+2) = 0
0(2) = 0
0 = 0 


Plug in x = -2
x(x+2) = 0
-2(-2+2) = 0
-2(0) = 0
0 = 0


So that shows how x = 0 or x = -2 makes the denominator x(x+2) equal to zero.


In set builder notation, the domain would be *[Tex \Large \{x|x\mathbb{R}, \ x \neq 0, \ x \neq -2\}]


In interval notation, the domain would be *[Tex \Large \left(-\infty, -2\right)\cup\left(-2, 0\right)\cup\left(0, \infty\right)]


Effectively we're "gluing" three intervals together. Or put another way, we're poking holes in the number line at -2 and 0 while leaving any other number as part of the domain. 
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