What is the extreme f(x)= x/(ln(x))[2.1] ?

The extreme f(x)= x/(ln(x))[2.1] is Minimum(e,5.70839…)

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extreme f(x)= x/(ln(x))[2.1]

extreme points

f (x) = \frac{x}{ln ( x )} [2.1]

Minimum (e, 5.70839 \dots)

Solution steps

f^{'} (x) = \frac{2.1 ( ln ( x ) - 1 )}{ln ^{2} ( x )}

Find intervals:

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < e,

Increasing

: e < x < \infty

Plug

x = e

into

\frac{x}{ln ( x )} [2.1] : 5.70839 \dots

Minimum (e, 5.70839 \dots)

f (x) = x^{4} - 72 x^{2} - 2, - 7 \leq x \leq 7

f (x) = x^{2} + 10 x^{\frac{2}{3}}

y = - x^{2} + 1 -, 1 \leq x \leq 2

f (x) = 3 x^{4} - 12 x^{3} - 60 x^{2} + 4

f (x) = 13 e^{- x}

What is the extreme f(x)= x/(ln(x))[2.1] ?
The extreme f(x)= x/(ln(x))[2.1] is Minimum(e,5.70839…)