What is the extreme e^{9x}+e^{-x} ?

The extreme e^{9x}+e^{-x} is Minimum(-(ln(3))/5 , 1/(e^{(9ln(3)){5}}}+e^{\frac{ln(3))/5})

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extreme e^{9x}+e^{-x}

extreme points

e^{9 x} + e^{- x}

Minimum (- \frac{ln ( 3 )}{5}, \frac{1}{e ^{\frac{9 l n ( 3 )}{5}}} + e^{\frac{l n ( 3 )}{5}})

Solution steps

f^{'} (x) = e^{9 x} \cdot 9 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{5} ln (3),

Increasing

: - \frac{1}{5} ln (3) < x < \infty

Plug

x = - \frac{ln ( 3 )}{5}

into

e^{9 x} + e^{- x} : \frac{1}{e ^{\frac{9 l n ( 3 )}{5}}} + e^{\frac{l n ( 3 )}{5}}

Minimum (- \frac{ln ( 3 )}{5}, \frac{1}{e ^{\frac{9 l n ( 3 )}{5}}} + e^{\frac{l n ( 3 )}{5}})

f (x) = x^{3} - 3 x y^{2} - 3 y^{2} + 12 y

x^{2} y - 6 y^{2} - 3 x^{2}

f (x) = - x^{2} + 7 x - 7

f (x) = x^{3} + x^{2} - 46 x + 80

f (x) = 3 x^{4} + 8 x^{3} - 6 x^{2} + 24 x

What is the extreme e^{9x}+e^{-x} ?
The extreme e^{9x}+e^{-x} is Minimum(-(ln(3))/5 , 1/(e^{(9ln(3)){5}}}+e^{\frac{ln(3))/5})