document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=85113>Question 85113</A>: <I><SPAN STYLE=\"word-break: break-all;\"> im really not sure how to do this i have to find the domain and range of y=square root of 3+x. how do i find the domain and range? thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #61328 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=61328\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> find the domain and range of y=square root of 3+x. how do i find the domain and range?<BR>\n" );
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document.write( "To find the domain you have to look for Real Number Values that x CANNOT take.<BR>\n" );
document.write( "In your problem you have y=sqrt(3+x)<BR>\n" );
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document.write( "That 3+x cannot be negative;it must be >=0<BR>\n" );
document.write( "So solve 3+x>=0<BR>\n" );
document.write( "x>-3<BR>\n" );
document.write( "The domain is all Real Numbers >=3<BR>\n" );
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document.write( "For the range you need to determine the Real Number values y CAN take.<BR>\n" );
document.write( "Since you have y=sqrt(3+x), y must be >=0<BR>\n" );
document.write( "The range is All Real Numbers >=0<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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