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

The extreme f(x)=5x-4ln(4x) is Minimum(4/5 ,4-4ln(16/5))

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

extreme points

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

Minimum (\frac{4}{5}, 4 - 4 ln (\frac{16}{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 (4 x) : 4 - 4 ln (\frac{16}{5})

Minimum (\frac{4}{5}, 4 - 4 ln (\frac{16}{5}))

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

g (x, y) = x^{3} + y^{3} - 3 x - 3 y

f (x) = - x^{6} - 3 x^{3} + x + 1

x^{2} = - 4 a y

f (x) = 3 sin (x) cos (x)

What is the extreme f(x)=5x-4ln(4x) ?
The extreme f(x)=5x-4ln(4x) is Minimum(4/5 ,4-4ln(16/5))