f = x^x = e^(xln(x))
f' = e^(xln(x))(ln(x)+1)
f' = x^x(ln(x)+1)
I'll let you take it from here
Hello,
I need help with the analysis of the following function.
Before to draw it need to find intersection points, min. max. asymptotes, discuss the domain, convexity/concavity by the usage of the second derivative test.. Having difficuloties with the following function:
f= x^x
and
f=e^(-x^2)
any help is welcome.
Thanks,
x^x is only defined for x> 0 so it makes no sense in talking asbout the limit
as x goes to - infinity since x is never negative
You would have to know L Hopital's rule in order to show x^x->1 as
x ->0 :
write y = lim (x^x)
x ->0
lny = lim xln(x)
x->0
lim xln(x) = lim ln(x)/(1/x) = lim -x = 0
lny = 0
y = e^0 = 1
i.e lim x^x = 1 as x goes to 0
Without L'hopital's rule you'll have to rely on a graph