What is the extreme e^{8x}+e^{-x} ?

The extreme e^{8x}+e^{-x} is Minimum(-(ln(2))/3 , 1/(e^{(8ln(2)){3}}}+e^{\frac{ln(2))/3})

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extreme e^{8x}+e^{-x}

extreme points

e^{8 x} + e^{- x}

Minimum (- \frac{ln ( 2 )}{3}, \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}})

Solution steps

f^{'} (x) = e^{8 x} \cdot 8 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{3} ln (2),

Increasing

: - \frac{1}{3} ln (2) < x < \infty

Plug

x = - \frac{ln ( 2 )}{3}

into

e^{8 x} + e^{- x} : \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}}

Minimum (- \frac{ln ( 2 )}{3}, \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}})

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

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

f (t) = 10 e^{a t}

f (x) = 2 x^{3} - 6 x^{2} - 90 x + 1

f (x) = (11 - x) (x + 1)^{2}

What is the extreme e^{8x}+e^{-x} ?
The extreme e^{8x}+e^{-x} is Minimum(-(ln(2))/3 , 1/(e^{(8ln(2)){3}}}+e^{\frac{ln(2))/3})