Calculus of a Single Variable, Chapter 5, 5.1, Section 5.1, Problem 80

We are asked to find the relative extrema and inflection points for the graph of y=2x-ln(2x) :
Note that the domain for the function is x>0.
Extrema can only occur at critical points; i.e. when the first derivative is zero or fails to exist.
y'=2-2/(2x) ==> y'=2-1/x
The first derivative exists for all values of x in the domain:
2-1/x=0==> 2=1/x ==> x=1/2
For 01/2 it is positive so there is a minimum at x=1/2. This is the only max or min.
Inflection points can only occur when the second derivative is zero:
y''=1/x^2>0 forall x so there are no inflection points. (The graph is concave up everywhere.)
The graph: