What is the extreme f(x)=x+e^{-4x} ?

The extreme f(x)=x+e^{-4x} is Minimum((ln(2))/2 ,(ln(2))/2+1/4)

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extreme f(x)=x+e^{-4x}

extreme points

f (x) = x + e^{- 4 x}

Minimum (\frac{ln ( 2 )}{2}, \frac{ln ( 2 )}{2} + \frac{1}{4})

Solution steps

f^{'} (x) = 1 - 4 e^{- 4 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^{- 4 x} : \frac{ln ( 2 )}{2} + \frac{1}{4}

Minimum (\frac{ln ( 2 )}{2}, \frac{ln ( 2 )}{2} + \frac{1}{4})

f (x) = - 2 x^{3} + 9 x^{2} - 12 x

- \frac{5 x}{x ^{2} + 4}

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

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

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

What is the extreme f(x)=x+e^{-4x} ?
The extreme f(x)=x+e^{-4x} is Minimum((ln(2))/2 ,(ln(2))/2+1/4)