What is the extreme f(x)=((3+9ln(x)))/x ?

The extreme f(x)=((3+9ln(x)))/x is Maximum(e^{2/3}, 9/(e^{2/3)})

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extreme f(x)=((3+9ln(x)))/x

extreme points

f (x) = \frac{( 3 + 9 ln ( x ) )}{x}

Maximum (e^{\frac{2}{3}}, \frac{9}{e ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = \frac{- 9 ln ( x ) + 6}{x ^{2}}

Find intervals:

Increasing

: 0 < x < e^{\frac{2}{3}},

Decreasing

: e^{\frac{2}{3}} < x < \infty

Plug

x = e^{\frac{2}{3}}

into

\frac{3 + 9 ln ( x )}{x} : \frac{9}{e ^{\frac{2}{3}}}

Maximum (e^{\frac{2}{3}}, \frac{9}{e ^{\frac{2}{3}}})

\frac{1}{3 x}

x + 3

f (x) = 4 x^{3} - 48 x - 5

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

f (x) = \frac{1}{x x ^{2} + 2}

What is the extreme f(x)=((3+9ln(x)))/x ?
The extreme f(x)=((3+9ln(x)))/x is Maximum(e^{2/3}, 9/(e^{2/3)})