What is the extreme f(x)=x+e^{-2x} ?

The extreme f(x)=x+e^{-2x} is Minimum((ln(2))/2 ,(ln(2)+1)/2)

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extreme f(x)=x+e^{-2x}

extreme points

f (x) = x + e^{- 2 x}

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{ln ( 2 )}{2},

Increasing

: \frac{ln ( 2 )}{2} < x < \infty

Plug

x = \frac{ln ( 2 )}{2}

into

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

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

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

f (x) = - x (x + 2) (x - 2)

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 5

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

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

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