Question 1035333
{{{f (x) = (lnx)^2 - 3lnx }}}  ==>  {{{df(x)/dx = (2lnx-3)/x}}}.

Now find the critical points of the function by letting {{{(2lnx-3)/x = 0}}}

==> {{{2lnx - 3=0}}}, or {{{x = e^(3/2)}}} or {{{x = e^1.5}}}.

Now apply the first derivative test:

When {{{x > e^1.5}}}, {{{df(x)/dx > 0}}} ==> f(x) is increasing over there.
When {{{x < e^1.5}}}, {{{df(x)/dx < 0}}} ==> f(x) is decreasing over there.

Thus there is a local minimum at {{{x = e^1.5}}}, and since it is the only critical point in an open interval (0, {{{infinity}}}), it is also an absolute minimum.