Find the absolute exterma of f(x) = x^x on (0,∞)?

Answer 1

x^x = e^xlnx and
y´= e^xlnx(lnx+1)= 0 at x=1/e and f(1/e) =
(1/e)^(1/e)
The sign of the derivative is ------1/e++++++
At 0+ lim f(x) with x=>0+=1 as xlnx=>0(you can verify it using L´Hôpital for lnx/(1/x))
f(1/e) is a relative minimum and also the absolute minimum.
As the function is increasing for x>1/e there is no absolute maximum

Answer 2

The extremum of f(x) is y = 0 at x = 0.