What is the extreme f(x)=ln((4x^2)/(ln(x))) ?

The extreme f(x)=ln((4x^2)/(ln(x))) is Minimum(sqrt(e),ln(8e))

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extreme f(x)=ln((4x^2)/(ln(x)))

extreme points

f (x) = ln (\frac{4 x ^{2}}{ln ( x )})

Minimum (e, ln (8 e))

Solution steps

f^{'} (x) = \frac{2 ln ( x ) - 1}{x ln ( x )}

Find intervals:

Decreasing

: 1 < x < e,

Increasing

: e < x < \infty

Plug

x = e

into

ln (\frac{4 x ^{2}}{ln ( x )}) : ln (8 e)

Minimum (e, ln (8 e))

f (x) = - x^{3} + 3 x^{2} + 9 x - 27

f (x) = 2 x^{3} + 3 x^{2} - 2 x - 3

\frac{( x + 1 )}{x ^{2} - 2 x - 3}

f (x) = \frac{x ^{2} + 2}{x ^{2} - 9}

f (x) = \frac{x ^{3}}{3} - 5 x^{2} - 50 x + 1

What is the extreme f(x)=ln((4x^2)/(ln(x))) ?
The extreme f(x)=ln((4x^2)/(ln(x))) is Minimum(sqrt(e),ln(8e))