What is the extreme f(x)=(ln(x))/(x^{5/2)} ?

The extreme f(x)=(ln(x))/(x^{5/2)} is Maximum(e^{2/5}, 2/(5e))

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extreme f(x)=(ln(x))/(x^{5/2)}

extreme

Maximum (e^{\frac{2}{5}}, \frac{2}{5 e})

Solution steps

f^{'} (x) = \frac{2 - 5 ln ( x )}{2 x ^{\frac{7}{2}}}

Find intervals:

Increasing

: 0 < x < e^{\frac{2}{5}},

Decreasing

: e^{\frac{2}{5}} < x < \infty

Plug

x = e^{\frac{2}{5}}

into

\frac{ln ( x )}{x ^{\frac{5}{2}}} : \frac{2}{5 e}

Maximum (e^{\frac{2}{5}}, \frac{2}{5 e})

What is the extreme f(x)=(ln(x))/(x^{5/2)} ?
The extreme f(x)=(ln(x))/(x^{5/2)} is Maximum(e^{2/5}, 2/(5e))