document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=390517>Question 390517</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the number which when raised by itself, it will become minimum. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #276916 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=276916\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+x%5Ex ALIGN=MIDDLE  ALT=\"y+=+x%5Ex\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  We have to restrict the domain of y to (0, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=infinity ALIGN=MIDDLE  ALT=\"infinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >), otherwise we encounter a densely infinite number of problems, haha.(E.g., <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%2F4%29%5E%28-1%2F4%29+=+1%2F%28%28-1%2F4%29%5E%281%2F4%29%29 ALIGN=MIDDLE  ALT=\"%28-1%2F4%29%5E%28-1%2F4%29+=+1%2F%28%28-1%2F4%29%5E%281%2F4%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-pi%29%5E%28-pi%29+=+1%2F%28%28-pi%29%5Epi%29 ALIGN=MIDDLE  ALT=\"%28-pi%29%5E%28-pi%29+=+1%2F%28%28-pi%29%5Epi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >).  Note also that y > 0 for all x in the domain.\r<BR>\n" );
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document.write( "That said, we can take ln of both sides of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+x%5Ex ALIGN=MIDDLE  ALT=\"y+=+x%5Ex\"   DISABLED_style= \"max-width:90%; height:auto;\" >:<BR>\n" );
document.write( "ln y  = x lnx.<BR>\n" );
document.write( "Differentiating both sides:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28dy%2Fdx%29%2Fy+=+lnx+%2B1 ALIGN=MIDDLE  ALT=\"%28dy%2Fdx%29%2Fy+=+lnx+%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" > <-----(A)<BR>\n" );
document.write( "Set <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dy%2Fdx+=+0 ALIGN=MIDDLE  ALT=\"dy%2Fdx+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so that the equation (A) becomes lnx = -1, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+e%5E%28-1%29+=+1%2Fe ALIGN=MIDDLE  ALT=\"x+=+e%5E%28-1%29+=+1%2Fe\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Since there is only 1 critical value this is either an absolute min or absolute max.  We find out using the 2nd derivative test.\r<BR>\n" );
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document.write( "Differentiate (A) again:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y%28d%5E2y%2Fdx%5E2%29+-+%28dy%2Fdx%29%5E2%29%2Fy%5E2+=+1%2Fx ALIGN=MIDDLE  ALT=\"%28y%28d%5E2y%2Fdx%5E2%29+-+%28dy%2Fdx%29%5E2%29%2Fy%5E2+=+1%2Fx\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "When x = 1/e, the equation becomes <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28d%5E2y%2Fdx%5E2%29%2Fy+=+e ALIGN=MIDDLE  ALT=\"%28d%5E2y%2Fdx%5E2%29%2Fy+=+e\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d%5E2y%2Fdx%5E2+=+ye+%3E+0 ALIGN=MIDDLE  ALT=\"d%5E2y%2Fdx%5E2+=+ye+%3E+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Thus the graph of the function y is concave up at x=1/e, and so there is an absolute minimum there for y. </SPAN>\n" );
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