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

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

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

extreme points

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

Minimum (e, 2 e)

Solution steps

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

Find intervals:

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < e,

Increasing

: e < x < \infty

Plug

x = e

into

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

Minimum (e, 2 e)

f (x, y) = x^{6} + x y - 2 y - x + 3

f (x) = 3 cos^{2} (x)

f (x) = 6 x - 7 x^{\frac{6}{7}}

- x^{3}

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

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