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 0<x<1/2 the first derivative is negative, for x>1/2 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:

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