extreme e^{2x}+e^{-x}

Lösung

extrempunkte

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^{2} + 3 x y + 4 y^{2} + 7 x + 11 y

e x t re m e f (x) = x^{3} - 3 x^{2} - 9 x + 8

e x t re m e f (x) = 25 x^{2} e^{2 x} - 16

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

e x t re m e y = x^{3} - 6 x^{2} + 9