What is the extreme f(x)=e^{4x}+e^{-x} ?

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

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extreme f(x)=e^{4x}+e^{-x}

extreme points

f (x) = 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, y) = 3 x^{2} + 2 y^{2} - 18 x + 16 y + 7

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

y = \frac{- 3 x}{x ^{2} + 2}

f (x) = 2 π r^{2} + \frac{1500}{r}

(x^{2} + 10) (4 - x^{2})

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