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		<title><![CDATA[Graph Forums — Evil math problem]]></title>
		<link>https://forum.padowan.dk/viewtopic.php?id=772</link>
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		<description><![CDATA[The most recent posts in Evil math problem.]]></description>
		<lastBuildDate>Mon, 14 Jan 2013 04:23:50 +0000</lastBuildDate>
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			<title><![CDATA[Evil math problem]]></title>
			<link>https://forum.padowan.dk/viewtopic.php?pid=2107#p2107</link>
			<description><![CDATA[<p>When graph is better than doing derivatives by hand...</p><p>find the inflection points of x^(x^e/e^x)</p><p>that is, take the second derivative and find the zeros.</p><p>or, in other words...</p><p>(e*x^(e-1)*e^x-x^e*e^x)/(e^x)^2*x^((x^e-e^x)/e^x)+x^e/e^x*((x^e-e^x)/e^x*x^((x^e-e^x-e^x)/e^x)+x^((x^e-e^x)/e^x)*ln(x)*((e*x^(e-1)-e^x)*e^x-(x^e-e^x)*e^x)/(e^x)^2)+((((x^e/e^x*x^((x^e-e^x)/e^x)+x^(x^e/e^x)*ln(x)*(e*x^(e-1)*e^x-x^e*e^x)/(e^x)^2)*ln(x)+x^(x^e/e^x)*1/x)*(e*x^(e-1)*e^x-x^e*e^x)+x^(x^e/e^x)*ln(x)*(e*(e-1)*x^(e-2)*e^x+e*x^(e-1)*e^x-(e*x^(e-1)*e^x+x^e*e^x)))*(e^x)^2-2*x^(x^e/e^x)*ln(x)*(e*x^(e-1)*e^x-x^e*e^x)*e^x*e^x)/((e^x)^2)^2</p><p>makes a great extra credit problem - it looks a little challenging but not THAT bad.</p>]]></description>
			<author><![CDATA[null@example.com (KFW)]]></author>
			<pubDate>Mon, 14 Jan 2013 04:23:50 +0000</pubDate>
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