Calculus of a Single Variable, Chapter 5, 5.1, Section 5.1, Problem 82

Locate extrema and inflection points for y=(ln(x))/x :
The domain is x>0.
Extrema can only occur at critical points, or points where the first derivative is zero or fails to exist.
y=1/x*ln(x)
y'=1/x*(1/x)+(-1/x^2)lnx
y'=1/x^2(1-lnx)
Setting equal to zero we get:
1/x^2=(lnx)/x^2 ==> lnx=1 ==> x=e
For 0e it is negative, so there is a maximum at x=e and this is the only extrema.
Inflection points can only occur if the second derivative is zero:
y''=-2/(x^3)(1-lnx)+1/x^2(-1/x)
y''=-1/x^3(2-2lnx+1)=-1/x^3(3-2lnx)
So 3-2lnx=0==> lnx=3/2 ==> x=e^(3/2)~~4.482
There is an inflection point at x=e^(1.5) where the graph changes from concave down to concave up.
The graph: