document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=878939>Question 878939</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Findind the domain of the rational function <BR>\n" );
document.write( "f(x)= 49x2-16<BR>\n" );
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document.write( "       2x-6 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #530387 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=530387\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You cannot divide by zero. If the denominator were zero, then \r<BR>\n" );
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document.write( "2x-6 = 0\r<BR>\n" );
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document.write( "2x = 6\r<BR>\n" );
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document.write( "x = 6/2\r<BR>\n" );
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document.write( "x = 3\r<BR>\n" );
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document.write( "That means if x = 3, then the denominator 2x-6 would be equal to zero.\r<BR>\n" );
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document.write( "So x = 3 causes a division by zero error. That forces us to exclude x = 3 from the domain.\r<BR>\n" );
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document.write( "Domain: the set of all real numbers x but x cannot equal 3\r<BR>\n" );
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document.write( "Another way to state the domain\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src=\"http://www.sciweavers.org/tex2img.php?eq=%5Ctext%7BDomain%3A%20%7D%5Cleft%5C%7Bx%7Cx%5Cin%5Cmathbb%7BR%7D%2Cx%20%5Cneq%203%5Cright%5C%7D&bc=White&fc=Black&im=jpg&fs=12&ff=arev&edit=0\" align=\"center\" border=\"0\" alt=\"\text{Domain: }\left\{x|x\in\mathbb{R},x \neq 3\right\}\" width=\"268\" height=\"18\" />\r<BR>\n" );
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document.write( "Domain in interval notation:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src=\"http://www.sciweavers.org/tex2img.php?eq=%5Cleft%28-%5Cinfty%2C3%5Cright%29%20%5Ccup%20%5Cleft%283%2C%5Cinfty%5Cright%29&bc=White&fc=Black&im=jpg&fs=12&ff=arev&edit=0\" align=\"center\" border=\"0\" alt=\"\left(-\infty,3\right) \cup \left(3,\infty\right)\" width=\"136\" height=\"18\" /> </SPAN>\n" );
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