What is the extreme f(x)=4ln(2xe^{-x}) ?

The extreme f(x)=4ln(2xe^{-x}) is Maximum(1,4(ln(2)-1))

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extreme f(x)=4ln(2xe^{-x})

extreme points

f (x) = 4 ln (2 x e^{- x})

Maximum (1, 4 (ln (2) - 1))

Solution steps

f^{'} (x) = 4 (\frac{1}{x} - 1)

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

4 ln (2 x e^{- x}) : 4 (ln (2) - 1)

Maximum (1, 4 (ln (2) - 1))

f (x) = 2 x^{3} - 3 x + 6

f (x) = \frac{y - 3}{y ^{2} - 3 y + 9}

f (x) = (16 - x) (100 x + 1300)

f (x) = \frac{1}{3 x ^{\frac{2}{3}}}

\frac{2 x - 3}{x}

What is the extreme f(x)=4ln(2xe^{-x}) ?
The extreme f(x)=4ln(2xe^{-x}) is Maximum(1,4(ln(2)-1))