document.write( "Question 383763: find the domain of the function f (x) = square root of 12 − 3x^2 \n" ); document.write( "
Algebra.Com's Answer #271759 by Edwin McCravy(20059) $\"\"$ $\"About$
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document.write( "The domain is the set of values of x that can be substituted into a \r\n" );
document.write( "functional equation.  \r\n" );
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document.write( "Negative numbers cannot be permitted under a square root radical, because you\r\n" );
document.write( "can't multiply a negative by itself, which is a negative) and get a negative,\r\n" );
document.write( "because you'll always get a positive,\r\n" );
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document.write( "So what is under that radical cannot be negative. So we cannot substitute any\r\n" );
document.write( "number for x and get an answer that will make  a negative number.\r\n" );
document.write( "So we set it  zero, so it won't be negative.\r\n" );
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document.write( "     \r\n" );
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document.write( "We can divide through both sides by 3 without reversing the inequality because\r\n" );
document.write( "we are dividing through by a positive number, not by a negative number.\r\n" );
document.write( "(If we had divided both sides by a negative number we would\r\n" );
document.write( "have had to reverse the symbol of imequality).\r\n" );
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document.write( "Now factor:\r\n" );
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document.write( "The critical values are found by setting the left side equal to zero.\r\n" );
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document.write( "2-x=0 has solution x=2\r\n" );
document.write( "2+x=0 has solution x=-2\r\n" );
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document.write( "Therefore these are the critical numbers.  We mark them on a number\r\n" );
document.write( "line with open circles \"o\" (but we may close them later).\r\n" );
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document.write( "-------------o---------------o---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0  +1  +2  +3  +4\r\n" );
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document.write( "We pick a test value left of -2, say -3 since it's the easiest. We substitute\r\n" );
document.write( "-3 in   \r\n" );
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document.write( "to see if it's true or false.\r\n" );
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document.write( "That's false so we do not shade that part of the number line, so we still just\r\n" );
document.write( "have this:\r\n" );
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document.write( "-------------o---------------o---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0   1   2   3   4\r\n" );
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document.write( "Next we pick a test value between -2 and 2, say 0 since it's the easiest. We\r\n" );
document.write( "substitute 0 in   \r\n" );
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document.write( "to see if it's true or false.\r\n" );
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document.write( "That's true so we shade that part of the number line, and now we have\r\n" );
document.write( "this:\r\n" );
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document.write( "-------------o===============o---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0  +1  +2  +3  +4\r\n" );
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document.write( "Next we pick a test value right of +2, say +3 since it's the easiest.\r\n" );
document.write( "We substitute +3 in   \r\n" );
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document.write( "to see if it's true or false.\r\n" );
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document.write( "That's false so we do not shade that part of the number line, \r\n" );
document.write( "so we still just have this:\r\n" );
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document.write( "-------------o===============o---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0  +1  +2  +3  +4\r\n" );
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document.write( "Next we test the critical points themselves, -2 and +2\r\n" );
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document.write( "Testing -2\r\n" );
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document.write( "This is true so we darken the circle at -2\r\n" );
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document.write( "-------------@===============o---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0  +1  +2  +3  +4\r\n" );
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document.write( "Testing +2\r\n" );
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document.write( "This is true too so we darken the circle at +2\r\n" );
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document.write( "-------------@===============@---------\r\n" );
document.write( "-5  -4  -3  -2  -1   0  +1  +2  +3  +4\r\n" );
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document.write( "In interval notation this is [-2, 2]\r\n" );
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document.write( "Edwin
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