What is the extreme (2x-1)ln(2x-1) ?

The extreme (2x-1)ln(2x-1) is Minimum((1+e}{2e}, 1/e ln(\frac{1+e)/e-1))

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extreme (2x-1)ln(2x-1)

extreme points

(2 x - 1) ln (2 x - 1)

Minimum (\frac{1 + e}{2 e}, \frac{1}{e} ln (\frac{1 + e}{e} - 1))

Solution steps

f^{'} (x) = 2 (ln (2 x - 1) + 1)

Find intervals:

Decreasing

: \frac{1}{2} < x < \frac{1 + e}{2 e},

Increasing

: \frac{1 + e}{2 e} < x < \infty

Plug

x = \frac{1 + e}{2 e}

into

(2 x - 1) ln (2 x - 1) : \frac{1}{e} ln (\frac{1 + e}{e} - 1)

Minimum (\frac{1 + e}{2 e}, \frac{1}{e} ln (\frac{1 + e}{e} - 1))

(- 7, 2), (9, - 9)

f (x) = \frac{2 x ^{3} + 3 x + 3}{5 x ^{3} + 4 x - 3}

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

f (x) = lo g_{3} (x)

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

What is the extreme (2x-1)ln(2x-1) ?
The extreme (2x-1)ln(2x-1) is Minimum((1+e}{2e}, 1/e ln(\frac{1+e)/e-1))