#### Topic: Evil math problem

When graph is better than doing derivatives by hand...

find the inflection points of x^(x^e/e^x)

that is, take the second derivative and find the zeros.

or, in other words...

(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

makes a great extra credit problem - it looks a little challenging but not THAT bad.