document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=883678>Question 883678</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The curve y = e^x + 4e^-2x has one stationary point.\r<BR>\n" );
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document.write( "Find the x-coordinate of this point<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #533691 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=533691\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We find the stationary point by finding dy/dx and setting the result to 0 and solving for x<BR>\n" );
document.write( "note that dy/dx of e^x is e^x and dy/dx of e^-2x is e^-2x, therefore<BR>\n" );
document.write( "dy/dx of e^x + 4e^-2x is e^x -8e^-2x<BR>\n" );
document.write( "now let's set it = 0<BR>\n" );
document.write( "e^x -8e^-2x = 0 then<BR>\n" );
document.write( "e*x = 8/e^2x<BR>\n" );
document.write( "multiply both sides of = by e^2x<BR>\n" );
document.write( "e^3x = 8<BR>\n" );
document.write( "use definition of logarithms<BR>\n" );
document.write( "3x = ln 8<BR>\n" );
document.write( "3x = 2.07944154<BR>\n" );
document.write( "x = 0.69314718<BR>\n" );
document.write( "this is the x co-ordinate of the stationary point\r<BR>\n" );
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