Question 383763
<pre>
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,

{{{"f(x)"}}}{{{""=""}}}{{{sqrt(12-3x^2)}}}

So what is under that radical cannot be negative. So we cannot substitute any
number for x and get an answer that will make {{{12-3x^2}}} a negative number.
So we set it {{{"">=""}}} zero, so it won't be negative.

{{{12-3x^2}}}{{{"">=""}}}{{{0}}}     

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.
<i><font size = 1>(If we had divided both sides by a negative number we would
have had to reverse the symbol of imequality)</i>.<font>

{{{12/3-3x^2/3}}}{{{"">=""}}}{{{0/3}}}

{{{4-x^2}}}{{{"">=""}}}{{{0}}}

Now factor:

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

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   

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

to see if it's true or false.

{{{(2-(-3))(2+(-3))}}}{{{"">=""}}}{{{0}}}
{{{(2+3)(2-3)}}}{{{"">=""}}}{{{0}}}
{{{(5)(-1)}}}{{{"">=""}}}{{{0}}}
{{{-5}}}{{{"">=""}}}{{{0}}}

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   

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

to see if it's true or false.

{{{(2-(0))(2+(0))}}}{{{"">=""}}}{{{0}}}
{{{(2)(2)}}}{{{"">=""}}}{{{0}}}
{{{4}}}{{{"">=""}}}{{{0}}}

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   

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

to see if it's true or false.

{{{(2-(+3))(2+(+3))}}}{{{"">=""}}}{{{0}}}
{{{(2-3)(5)}}}{{{"">=""}}}{{{0}}}
{{{(-1)(5)}}}{{{"">=""}}}{{{0}}}
{{{-5}}}{{{"">=""}}}{{{0}}}

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

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

{{{(2-(-2))(2+(-2))}}}{{{"">=""}}}{{{0}}}
{{{(2+2)(0)}}}{{{"">=""}}}{{{0}}}
{{{(4)(0)}}}{{{"">=""}}}{{{0}}}
{{{0}}}{{{"">=""}}}{{{0}}}

This is true so we darken the circle at -2

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

Testing +2

{{{(2-x)(2+x)}}}{{{"">=""}}}{{{0}}}

{{{(2-(+2))(2+(+2))}}}{{{"">=""}}}{{{0}}}
{{{(2-2)(4)}}}{{{"">=""}}}{{{0}}}
{{{(0)(4)}}}{{{"">=""}}}{{{0}}}
{{{0}}}{{{"">=""}}}{{{0}}}

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</pre>