SOLUTION: find the domain of the function f (x) = square root of 12 − 3x^2

The domain is the set of values of x that can be substituted into a 
functional equation.  

Negative numbers cannot be permitted under a square root radical, because you
can't multiply a negative by itself, which is a negative) and get a negative,
because you'll always get a positive,



So what is under that radical cannot be negative. So we cannot substitute any
number for x and get an answer that will make  a negative number.
So we set it  zero, so it won't be negative.

     

We can divide through both sides by 3 without reversing the inequality because
we are dividing through by a positive number, not by a negative number.
(If we had divided both sides by a negative number we would
have had to reverse the symbol of imequality).





Now factor:



The critical values are found by setting the left side equal to zero.

2-x=0 has solution x=2
2+x=0 has solution x=-2

Therefore these are the critical numbers.  We mark them on a number
line with open circles "o" (but we may close them later).

-------------o---------------o---------
-5  -4  -3  -2  -1   0  +1  +2  +3  +4

We pick a test value left of -2, say -3 since it's the easiest. We substitute
-3 in   



to see if it's true or false.






That's false so we do not shade that part of the number line, so we still just
have this:

-------------o---------------o---------
-5  -4  -3  -2  -1   0   1   2   3   4
 

Next we pick a test value between -2 and 2, say 0 since it's the easiest. We
substitute 0 in   



to see if it's true or false.





That's true so we shade that part of the number line, and now we have
this:

-------------o===============o---------
-5  -4  -3  -2  -1   0  +1  +2  +3  +4



Next we pick a test value right of +2, say +3 since it's the easiest.
We substitute +3 in   



to see if it's true or false.






That's false so we do not shade that part of the number line, 
so we still just have this:

-------------o===============o---------
-5  -4  -3  -2  -1   0  +1  +2  +3  +4


Next we test the critical points themselves, -2 and +2

Testing -2








This is true so we darken the circle at -2

-------------@===============o---------
-5  -4  -3  -2  -1   0  +1  +2  +3  +4

Testing +2








This is true too so we darken the circle at +2

-------------@===============@---------
-5  -4  -3  -2  -1   0  +1  +2  +3  +4

In interval notation this is [-2, 2]

Edwin