extreme f(x)=e^{2x}+e^{-x}

Lösung

extrempunkte

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

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

Schritte zur Lösung

f^{'} (x) = e^{2 x} \cdot 2 - 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^{2 x} + e^{- x} : \frac{1}{e ^{\frac{2 l n ( 2 )}{3}}} + e^{\frac{l n ( 2 )}{3}}

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

e x t re m e f (x) = - 2 x^{3} + 6 x^{2} - 3

e x t re m e f (x, y) = x + 2 y

e x t re m e f (x) = x^{3} - 9 x^{2} + 15 x - 5

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

e x t re m e f (x, y) = 3 x - x^{3} - 2 y^{2} + y^{4}