Given:
f(x) = x^5lnxf(x)=x5lnx
we note that the function is defined only for x >0x>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]}