Question #3b185

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Answer 1 · 2016-11-17T13:05:27

See below.

Calling $f(x) = x-elog_ex$ it's minimum is located at

$(df)/(dx)=1-e/x=0$

or at $x = e$

At this point we have

$f(e)=0$

To qualify this stationary point we compute

$(d^2f)/(dx^2) =e/x^2$ and for $x=e$ we have

$(d^2f)/(dx^2)=e^(-1) > 0$ so $x=e$ represents a minimum of $f(x)$

The conclusion is:

$e log_ex le x$ for $x > 0$ and also $e log_ex$ and $x$ osculate at $x = e$ .

Note that for $x < 0$ , $log_ex$ is not defined as a real function of a real variable.

Attached a plot with $elog_ex$ (blue) and $x$ (red)

enter image source here