What is the extreme f(x)=5x-4ln(3x) ?

The extreme f(x)=5x-4ln(3x) is Minimum(4/5 ,4-4ln(12/5))

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extreme f(x)=5x-4ln(3x)

extreme points

f (x) = 5 x - 4 ln (3 x)

Minimum (\frac{4}{5}, 4 - 4 ln (\frac{12}{5}))

Solution steps

f^{'} (x) = 5 - \frac{4}{x}

Find intervals:

Decreasing

: 0 < x < \frac{4}{5},

Increasing

: \frac{4}{5} < x < \infty

Plug

x = \frac{4}{5}

into

5 x - 4 ln (3 x) : 4 - 4 ln (\frac{12}{5})

Minimum (\frac{4}{5}, 4 - 4 ln (\frac{12}{5}))

\frac{6}{x + 7}

g (x, y) = - 4 c x + 4 y x

f (x, y) = (x^{2} + y^{2})^{2} = 2 \cdot (x^{2} - y^{2})

f (x) = \frac{x ^{2}}{x ^{2} - 36}

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

What is the extreme f(x)=5x-4ln(3x) ?
The extreme f(x)=5x-4ln(3x) is Minimum(4/5 ,4-4ln(12/5))