What is the extreme f(x)=e^{3x}+e^{-x} ?

The extreme f(x)=e^{3x}+e^{-x} is Minimum(-(ln(3))/4 , 1/(e^{(3ln(3)){4}}}+e^{\frac{ln(3))/4})

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

extreme points

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

Minimum (- \frac{ln ( 3 )}{4}, \frac{1}{e ^{\frac{3 l n ( 3 )}{4}}} + e^{\frac{l n ( 3 )}{4}})

Solution steps

f^{'} (x) = e^{3 x} \cdot 3 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{4} ln (3),

Increasing

: - \frac{1}{4} ln (3) < x < \infty

Plug

x = - \frac{ln ( 3 )}{4}

into

e^{3 x} + e^{- x} : \frac{1}{e ^{\frac{3 l n ( 3 )}{4}}} + e^{\frac{l n ( 3 )}{4}}

Minimum (- \frac{ln ( 3 )}{4}, \frac{1}{e ^{\frac{3 l n ( 3 )}{4}}} + e^{\frac{l n ( 3 )}{4}})

1 - x

f (x) = \frac{1 - 5 x}{6 x}

\frac{1}{\frac{1}{x}}

f (x) = 2 x + 3 - \frac{1}{3} x^{2}

\frac{4}{x - 3}

What is the extreme f(x)=e^{3x}+e^{-x} ?
The extreme f(x)=e^{3x}+e^{-x} is Minimum(-(ln(3))/4 , 1/(e^{(3ln(3)){4}}}+e^{\frac{ln(3))/4})