Given:

#f(x) = x^5lnx#

we note that the function is defined only for #x >0# and that using l'Hospital's rule:

#lim_(x->0^+) x^5lnx = lim_(x->0^+) lnx/(1/x^5) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/x^5) = lim_(x->0^+) (1/x)(1/(-5/x^6)) = lim_(x->0^+) -x^5/5 = 0#

evaluate the derivative of the function:

#f'(x) = d/dx (x^5lnx) = d/dx(x^5)lnx + x^5 d/dx(lnx) = 5x^4lnx +x^5/x =x^4(5lnx+1)#

We can then find critical points for the function solving the equation:

#x^4(5lnx+1) = 0#

since #x > 0# the only solution is:

#5lnx+1 = 0 => x=e^(-1/5)#

Now we can evaluate the inequality:

#f'(x) > 0#

Since #x^4 >0# for #x>0# the sign of the derivative is the same as the sign of #5lnx+1#, which means:

#f'(x) > 0# for #x in (e^(-1/5), +oo)#

and #f'(x) < 0# for #x in (0,e^(-1/5))#

This shows that the function is strictly decreasing in #(0,e^(-1/5))# and strictly increasing in # (e^(-1/5), +oo)#, which means that:

#x= e^(-1/5) ~= 0.81873...#

is a local minimum.

graph{x^5lnx [-1.25, 1.25, -0.625, 0.625]}