What is the extreme e^{7x}+e^{-x} ?

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

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extreme e^{7x}+e^{-x}

extreme points

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, y) = 2 + 2 x + 4 y - x^{2} - y^{2}

f (x) = x 81 - x^{2}

S (a, d) = a + (n - 1) d

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

y = 3 x + 2 cos (x), 0 \leq x \leq 2 π

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