document.write( "Question 105759: Find the domain of y=1/(x+1)(x-3)
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Algebra.Com's Answer #76974 by jim_thompson5910(35256)\"\" \"About 
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\"y=%281%29%2F%28%28x-3%29%28x%2B1%29%29\" Start with the given function\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%28x%2B1%29=0\" Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.\r
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\n" ); document.write( "\n" ); document.write( "Now set each factor equal to zero:\r
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\n" ); document.write( "\n" ); document.write( "\"x-3=0\" or \"x%2B1=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=3\" or \"x=-1\" Now solve for x in each case\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are \"x=3\" or \"x=-1\"\r
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\n" ); document.write( "\n" ); document.write( "Since \"x=-1\" and \"x=3\" make the denominator equal to zero, this means we must exclude \"x=-1\" and \"x=3\" from our domain\r
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\n" ); document.write( "\n" ); document.write( "So our domain is: \r
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\n" ); document.write( "\n" ); document.write( "which in plain English reads: x is the set of all real numbers except \"x%3C%3E-1\" or \"x%3C%3E3\"\r
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\n" ); document.write( "\n" ); document.write( "So our domain looks like this in interval notation\r
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\n" ); document.write( "\n" ); document.write( "note: remember, the parenthesis excludes -1 and 3 from the domain\r
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\n" ); document.write( "\n" ); document.write( "If we wanted to graph the domain on a number line, we would get:\r
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\n" ); document.write( "\n" ); document.write( " Graph of the domain in blue and the excluded values represented by open circles\r
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\n" ); document.write( "\n" ); document.write( "Notice we have a continuous line until we get to the holes at \"x=-1\" and \"x=3\" (which is represented by the open circles).
\n" ); document.write( "This graphically represents our domain in which x can be any number except x cannot equal -1 or 3\r
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