What is the extreme f(x)=e^{6x}+e^{-x} ?

The extreme f(x)=e^{6x}+e^{-x} is Minimum(-(ln(6))/7 , 1/(e^{(6ln(6)){7}}}+e^{\frac{ln(6))/7})

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extreme f(x)=e^{6x}+e^{-x}

extreme points

f (x) = e^{6 x} + e^{- x}

Minimum (- \frac{ln ( 6 )}{7}, \frac{1}{e ^{\frac{6 l n ( 6 )}{7}}} + e^{\frac{l n ( 6 )}{7}})

Solution steps

f^{'} (x) = e^{6 x} \cdot 6 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{7} ln (6),

Increasing

: - \frac{1}{7} ln (6) < x < \infty

Plug

x = - \frac{ln ( 6 )}{7}

into

e^{6 x} + e^{- x} : \frac{1}{e ^{\frac{6 l n ( 6 )}{7}}} + e^{\frac{l n ( 6 )}{7}}

Minimum (- \frac{ln ( 6 )}{7}, \frac{1}{e ^{\frac{6 l n ( 6 )}{7}}} + e^{\frac{l n ( 6 )}{7}})

f (x) = \frac{x}{1 + x}

f (x) = x^{2} x + 5

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

f (x) = x^{2} - 8 x + 4

f (x) = e^{x^{2} - 5 x - 1}

What is the extreme f(x)=e^{6x}+e^{-x} ?
The extreme f(x)=e^{6x}+e^{-x} is Minimum(-(ln(6))/7 , 1/(e^{(6ln(6)){7}}}+e^{\frac{ln(6))/7})