document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1100669>Question 1100669</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the open intervals where the function is concave up and concave down.<BR>\n" );
document.write( "f(x)= x^3 -3x^2 -1\r<BR>\n" );
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document.write( "Answer: concave up (1,infinity) <BR>\n" );
document.write( "               concave down (-infinity, 1)\r<BR>\n" );
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document.write( "i dont understand because i got 0,2 for the critical points when you equal the whole derivative to zero. So please explain how to do it by hand without the calculator. Thanks. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #715201 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=715201\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f(x)= x^3 -3x^2 -1 <BR>\n" );
document.write( "It's true that the first derivative shows critical points of 0, 2<BR>\n" );
document.write( "The inflection point uses the second derivative, which is 6x-6<BR>\n" );
document.write( "at 0, f''(x) is -6 and at (0.5) it is -3 and at 1 it is 0 and at 1.5 it is +3<BR>\n" );
document.write( "Therefore, the concavity changes at 1.  The critical points will be where there are local maxima or minima; the concavity changes are in between somewhere. </SPAN>\n" );
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