`y = 2x - ln(2x)` Locate any relative extrema and points of inflection.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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:


Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial