What is the extreme f(x)=e^{5x}+e^{-x} ?

The extreme f(x)=e^{5x}+e^{-x} is Minimum(-(ln(5))/6 , 1/(e^{(5ln(5)){6}}}+e^{\frac{ln(5))/6})

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extreme f(x)=e^{5x}+e^{-x}

extreme points

f (x) = e^{5 x} + e^{- x}

Minimum (- \frac{ln ( 5 )}{6}, \frac{1}{e ^{\frac{5 l n ( 5 )}{6}}} + e^{\frac{l n ( 5 )}{6}})

Solution steps

f^{'} (x) = e^{5 x} \cdot 5 - e^{- x}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{6} ln (5),

Increasing

: - \frac{1}{6} ln (5) < x < \infty

Plug

x = - \frac{ln ( 5 )}{6}

into

e^{5 x} + e^{- x} : \frac{1}{e ^{\frac{5 l n ( 5 )}{6}}} + e^{\frac{l n ( 5 )}{6}}

Minimum (- \frac{ln ( 5 )}{6}, \frac{1}{e ^{\frac{5 l n ( 5 )}{6}}} + e^{\frac{l n ( 5 )}{6}})

f (x) = x^{2} + x y + y^{2} + 3 x - 3 y + 4

x^{5} - 15 x^{3}

f (x, y) = ln (4 x^{2} + 18 y^{2} - 36)

F (x, y) = x^{3} - 3 x y + 3 y^{2} + 1

f (x) = x y + 5 x - 5

What is the extreme f(x)=e^{5x}+e^{-x} ?
The extreme f(x)=e^{5x}+e^{-x} is Minimum(-(ln(5))/6 , 1/(e^{(5ln(5)){6}}}+e^{\frac{ln(5))/6})