Find where #f''(x)=0# OR #f''(x)# switches from negative to positive instantaneously (like with an asymptote).
First, we find points where #f''(x)=0#.
#f'(x)=(lnx-x(1/x))/(lnx)^2=1/(lnx)-1/(lnx)^2#
#f''(x) = -1/(x(lnx)^2) +2/(x(lnx)^3)#
Now set this equal to 0.
#0 = -1/(x(lnx)^2) +2/(x(lnx)^3)#
#0 = -lnx+2#
#lnx = 2#
#x = e^2#
This shows that the only solution to #f''(x)=0# is #x=e^2#. Additionally, in order to be a point of inflection, the graph of #f''(x)# must cross the x-axis from positive to negative or negative to positive AT #x=e^2# (which can be proven by showing that #d/dx(x/lnx)!=0# when #x=e^2#, but I will not include this proof unless it is requested).
So, #x=e^2# is a point of inflection.
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We also need to look for points of instantaneous change. In this case, we have one: the asymptote at #x=1#.
At #x=1#, #ln(x)=0#, and it switches from negative to positive by definition. This means that #-1/(x(lnx)^2) +2/(x(lnx)^3)# will also jump from negative to positive, since the denominator is switching signs. Even though #f''(x)# never equals 0, it switches signs at #x=1#, so #x=1# is a point of inflection.
Therefore the two points of inflection for #f(x)=x/lnx# are #x=1# and #x=e^2#
