We start by observing that the function:
#f(x) = lnx/sqrt(x)#
is defined in the interval: #x in (0,+oo)#, that it is negative for #x<1#, positive for #x>1# and has a single zero for #x=1#
We can analyze the behavior at the limits of the domain:
#lim_(x->0^+) lnx/sqrt(x) = -oo#
#lim_(x->oo) lnx/sqrt(x) = 0#
so the function has the line#x=0# for vertical asymptote and the line #y=0# has horizontal asymptote.
Let's calculate the first and second derivative of the function:
#f'(x) = (sqrtx/x-lnx/(2sqrtx))/x = sqrt(x)/x^2 -lnx/(2xsqrtx) = (2-lnx)/(2xsqrtx)#
#f''(x) = ((-(2xsqrtx)/x)-((2-lnx)3sqrtx))/(4x^3) = (-2sqrtx-6sqrtx+3lnxsqrtx)/(4x^3)= (3lnx-8)/(4x^2sqrtx)#
Now we can see that:
(1) #f'(x) > 0# for #2-lnx >0# that is for #x < e^2#
so the function is increasing in the interval #(0,e^2)# and decreasing in #(e^2,+oo)# and has a single critical value in #x=e^2# that is a local maximum.
(2) #f''(x) > 0# for #3lnx-8 > 0 # or #x > e^(8/3)#
so the function is concave down in the interval #(0,e^(8/3))# and concave up in #(e^(8/3),+oo)# and has a single inflection point in #x=e^(8/3)#.
