What are the local extrema, if any, of $f (x) =(lnx)^2/x$ ?

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Answer 1 · 2016-01-06T16:37:51

There is a local minimum of $0$ at $1$ . (Which is also global.) and a local maximum of $4/e^2$ at $e^2$ .

For $f (x) =(lnx)^2/x$ , note first that the domain of $f$ is the positive real numbers, $(0,oo)$ .

Then find

$f'(x) = ([2(lnx)(1/x)]*x - (lnx)^2[1])/x^2$

$= (lnx(2-lnx))/x^2$ .

$f'$ is undefined at $x=0$ which is not in the domain of $f$ , so it is not a critical number for $f$ .

$f'(x)=0$ where

$lnx=0$ or $2-lnx=0$

$x=1$ or $x=e^2$

Test the intervals $(0,1)$ , $(1,e^2)$ , and $(e^2,oo)$ .

(For test numbers, I suggest $e^-1, e^1, e^3$ -- recall $1=e^0$ and $e^x$ is increasing.)

We find that $f'$ changes from negative to positive as we pass $1$ , so $f(1)=0$ is a local minimum,

and that $f'$ changes from positive to negative as we pass $e^2$ , so $f(e^2)=4/e^2$ is a local maximum.

What are the local extrema, if any, of f (x) =(lnx)^2/xf (x) =(lnx)^2/x?