extreme f(x)=e^{8x}+e^{-x}

Lösung

extrempunkte

f (x) = e^{8 x} + e^{- x}

Minimum (- \frac{ln ( 2 )}{3}, \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}})

Schritte zur Lösung

f^{'} (x) = e^{8 x} \cdot 8 - e^{- x}

Intervalle finden:

Absteigend

: - \infty < x < - \frac{1}{3} ln (2),

Ansteigend

: - \frac{1}{3} ln (2) < x < \infty

Stecke

x = - \frac{ln ( 2 )}{3}

e^{8 x} + e^{- x} : \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}}

Minimum (- \frac{ln ( 2 )}{3}, \frac{1}{e ^{\frac{8 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}})

e x t re m e f (x) = - 2 x^{2} + 4 x

e x t re m e g (x) = x^{3} - 3 x^{2} + 1

e x t re m e f (x) = \frac{( x - 1 ) ^{2}}{x}

e x t re m e f (x, y) = 4 x y + \frac{2}{x} + \frac{1}{y}

e x t re m e f (x) = x^{4} - 8 x^{3} + 6