document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1027710>Question 1027710</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider this function f(x) = x^3+ax^2+bx+c where a>=0, b>0<BR>\n" );
document.write( "Over what intervals is f concave up? Concave down?<BR>\n" );
document.write( "Show that f must have one inflection point.<BR>\n" );
document.write( "Given that (0,-2) is the inflection point of f, solve for a and c and then show that f has no critical points. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #642984 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=642984\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+x%5E3%2Bax%5E2%2Bbx%2Bc+ ALIGN=MIDDLE  ALT=\"f%28x%29+=+x%5E3%2Bax%5E2%2Bbx%2Bc+\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "==> f'(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B2ax+%2B+b ALIGN=MIDDLE  ALT=\"3x%5E2%2B2ax+%2B+b\"   DISABLED_style= \"max-width:90%; height:auto;\" > ==> f\"(x) = 6x + 2a\r<BR>\n" );
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document.write( "Setting f\" to zero, we get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+-a%2F3 ALIGN=MIDDLE  ALT=\"x+=+-a%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "To the right of -a/3, f\" >0, while to its left f\" < 0.  (a > 0!)\r<BR>\n" );
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document.write( "==> To the right of -a/3, f is concave up, while to its left f is concave down.\r<BR>\n" );
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document.write( "Hence <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+-a%2F3 ALIGN=MIDDLE  ALT=\"x+=+-a%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > is an inflection point of f.\r<BR>\n" );
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document.write( "If (0,-2) is an inflection point then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-a%2F3+=+0 ALIGN=MIDDLE  ALT=\"-a%2F3+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and so a = 0.\r<BR>\n" );
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document.write( "==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+x%5E3%2Bbx%2Bc+ ALIGN=MIDDLE  ALT=\"f%28x%29+=+x%5E3%2Bbx%2Bc+\"   DISABLED_style= \"max-width:90%; height:auto;\" > ==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%280%29+=+0%5E3%2Bb%2A0%2Bc++=+-2 ALIGN=MIDDLE  ALT=\"f%280%29+=+0%5E3%2Bb%2A0%2Bc++=+-2\"   DISABLED_style= \"max-width:90%; height:auto;\" > ==> c = -2\r<BR>\n" );
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document.write( "==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28f%28x%29+=+x%5E3%2Bbx-2%29+ ALIGN=MIDDLE  ALT=\"highlight%28f%28x%29+=+x%5E3%2Bbx-2%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "==> f'(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E2%2B+b ALIGN=MIDDLE  ALT=\"3x%5E2%2B+b\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Since f' > 0 for all x because b > 0, f will not have any critical points, and the problem is solved.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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