What is the extreme e^{4x}+e^{-x} ?

The extreme e^{4x}+e^{-x} is Minimum(-(2ln(2))/5 , 1/(e^{(8ln(2)){5}}}+e^{\frac{2ln(2))/5})

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extreme e^{4x}+e^{-x}

extreme points

e^{4 x} + e^{- x}

Minimum (- \frac{2 ln ( 2 )}{5}, \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}})

Solution steps

f^{'} (x) = e^{4 x} \cdot 4 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{2}{5} ln (2),

Increasing

: - \frac{2}{5} ln (2) < x < \infty

Plug

x = - \frac{2 ln ( 2 )}{5}

into

e^{4 x} + e^{- x} : \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}}

Minimum (- \frac{2 ln ( 2 )}{5}, \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}})

f (x) = x^{2} - 12 x + 8

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

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

12 x e^{- x}

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

What is the extreme e^{4x}+e^{-x} ?
The extreme e^{4x}+e^{-x} is Minimum(-(2ln(2))/5 , 1/(e^{(8ln(2)){5}}}+e^{\frac{2ln(2))/5})