document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=907423>Question 907423</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Using the intermediate value theorem, determine whether the function has at least one real zero between a and b. \r<BR>\n" );
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document.write( "    a. f(x) = x 3 + 3x 2 – 9x – 13       a = - 5    b = - 4\r<BR>\n" );
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document.write( "    b. f(x) = 3x 2 – 2x – 11      a = - 3     b = - 2\r<BR>\n" );
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document.write( "Thank you for the help in advance </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #550336 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=550336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Both problems are done the same way.  Calculate <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f(a)\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f(b)\">.  If the sign of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f(a)\"> is different than the sign of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f(b)\">, then for a function that is continuous over the given interval, there is at least one real zero in the interval.  If the signs are the same, the existence of a zero cannot be guaranteed, nor can the possibility be eliminated.  Both of your problems are polynomial functions so there are no continuity issues.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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