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

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

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

extreme points

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

Minimum (e^{2}, \frac{e ^{2}}{4})

Solution steps

f^{'} (x) = \frac{ln ( x ) - 2}{ln ^{3} ( x )}

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < e^{2},

Increasing

: e^{2} < x < \infty

Plug

x = e^{2}

into

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

Minimum (e^{2}, \frac{e ^{2}}{4})

f (x) = \frac{1}{x + 6}

(3.3) (- 3.1)

\frac{x ^{2} - 3 x - 4}{1 + 4 x + 4 x ^{2}}

f (x) = \frac{2}{x} + 2 x

\frac{8 x}{5} - 2 y = - 8

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