What are inflection points of f(x)=ln x/x, if x>0?

The inflection points are the roots of the second
derivative of the function.

We'll have to determine the 1st
derivative, for the beginning. We'll do this using quotient
rule:

f'(x) = [(ln x)'*x - (ln
x)*x']/x^2

f'(x) = (x/x - ln
x)/x^2

f'(x) = (1 - ln
x)/x^2

Now, we'll determine the 2nd derivative using
quotient rule:

f"(x) = [(1 - ln x)'*x^2 - (1 - ln
x)*(x^2)']/x^4

f"(x) = [(-1/x)*x^2 - 2x*(1 - ln
x)]/x^4

f"(x) = [-x - 2x + (2x*ln
x)]/x^4

f"(x) = (2ln x -
3)/x^3

We'll equate and we'll
get:

f"(x) = 0

(2ln x - 3)/x^3
= 0 <=> 2ln x - 3 = 0 => 2ln x = 3 => ln x =
3/2

x = e^(3/2)

x = sqrt
e^3

f(e^(3/2)) = [ln
e^(3/2)]/e^(3/2)

We'll use the power property of
logarithms:

f(e^(3/2)) =
3/2e^(3/2)

f(e^(3/2)) = 3/(2e*sqrt
e)

The inflection point has the coordinates:
(e^(3/2) ; 3/(2e*sqrt e)).