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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = e^{2 x} \cdot 2 - 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^{2 x} + e^{- x} : \frac{1}{e ^{\frac{2 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}}

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

- x^{2} + 8 x - 9

f (x) = x^{3} - 12 x^{2} - 27 x + 2, 1 \leq x \leq 10

85 y^{35}

2 x^{3} + 9 x^{2} - 24 x

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

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