Question 360500
{{{"f(x)"= sqrt(9-x^2)}}}
<pre>
We require that the expression under the radical, the "radicand",
is not negative, but is 0 or something greater. 

{{{9-x^2>=0}}}

{{{(3-x)(3+x)>=0}}}

The expression on the left has zeros 3 and -3, so we
mark these solid on a number line:

-------------@-----------------------@--------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5  

We choose a test value on the left of -3, say -4, and
substitute it in:

{{{(3-x)(3+x)>=0}}}
{{{(3-(-4))(3+(-4))>=0}}}
{{{(3+4)(3-4)>=0}}}
{{{(7)(-1)>=0}}}
{{{-7>=0}}}

This is false so we do not shade to the left of -3
So we still have:

-------------@-----------------------@--------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5

We choose a test value between -3 and +3, say 0, and
substitute it in:

{{{(3-x)(3+x)>=0}}}
{{{(3-(0))(3+(0))>=0}}}
{{{(3-0)(3+0)>=0}}}
{{{(3)(3)>=0}}}
{{{9>=0}}}

This is true so we shade the part of the number line between
-3 and +3 inclusive of -3 and 3 since the inequality is {{{"">=""}}}
and not >.  So we have:

-------------@=======================@--------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5

We choose a test value on the right of +3, say +4, and
substitute it in:

{{{(3-x)(3+x)>=0}}}
{{{(3-(4))(3+(4))>=0}}}
{{{(3-4)(3+4)>=0}}}
{{{(-1)(7)>=0}}}
{{{-7>=0}}}

This is false so we do not shade to the right of 3
So we still have:

-------------@=======================@--------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5

and the domain is [-3,3]

The integers in the domain are -3, -2, -1, 0, 1, 2, 3.

So the answer is 7, choice d.

Edwin</pre>