What is the extreme f(x)=e^{7x}+e^{-x} ?

The extreme f(x)=e^{7x}+e^{-x} is Minimum(-(ln(7))/8 , 1/(e^{(7ln(7)){8}}}+e^{\frac{ln(7))/8})

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extreme f(x)=e^{7x}+e^{-x}

extreme points

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

Minimum (- \frac{ln ( 7 )}{8}, \frac{1}{e ^{\frac{7 l n ( 7 )}{8}}} + e^{\frac{l n ( 7 )}{8}})

Solution steps

f^{'} (x) = e^{7 x} \cdot 7 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{8} ln (7),

Increasing

: - \frac{1}{8} ln (7) < x < \infty

Plug

x = - \frac{ln ( 7 )}{8}

into

e^{7 x} + e^{- x} : \frac{1}{e ^{\frac{7 l n ( 7 )}{8}}} + e^{\frac{l n ( 7 )}{8}}

Minimum (- \frac{ln ( 7 )}{8}, \frac{1}{e ^{\frac{7 l n ( 7 )}{8}}} + e^{\frac{l n ( 7 )}{8}})

f (x) = 8 x^{3} - x^{4} = x^{3} (8 - x)

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

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

f (x) = x^{3} - \frac{3}{2} x^{2}, - 3 \leq x \leq 6

f (x, y) = e^{2 x} - x^{3} y^{5}

What is the extreme f(x)=e^{7x}+e^{-x} ?
The extreme f(x)=e^{7x}+e^{-x} is Minimum(-(ln(7))/8 , 1/(e^{(7ln(7)){8}}}+e^{\frac{ln(7))/8})