What is the extreme e^{2x}+e^{-x} ?

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

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extreme e^{2x}+e^{-x}

extreme points

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}})

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

f (x) = 2 x^{2} + 3 x y + 4 y^{2} + 7 x + 11 y

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

f (x, y) = y - x + 2

f (x) = x^{3} - 3 x^{2} - 9 x + 8

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