What is the extreme f(x)=e^{8x}+e^{-x} ?

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

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extreme f(x)=e^{8x}+e^{-x}

extreme points

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

g (x) = x^{3} - 3 x^{2} + 1

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

f (x, y) = 4 x y + \frac{2}{x} + \frac{1}{y}

f (x) = x^{4} - 8 x^{3} + 6

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