extreme f(x)=e^{4x}+e^{-x}

Lösung

extrempunkte

f (x) = e^{4 x} + e^{- x}

Minimum (- \frac{2 ln ( 2 )}{5}, \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}})

Schritte zur Lösung

f^{'} (x) = e^{4 x} \cdot 4 - e^{- x}

Intervalle finden:

Absteigend

: - \infty < x < - \frac{2}{5} ln (2),

Ansteigend

: - \frac{2}{5} ln (2) < x < \infty

Stecke

x = - \frac{2 ln ( 2 )}{5}

e^{4 x} + e^{- x} : \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}}

Minimum (- \frac{2 ln ( 2 )}{5}, \frac{1}{e ^{\frac{8 l n ( 2 )}{5}}} + e^{\frac{2 l n ( 2 )}{5}})

e x t re m e y = \frac{- 3 x}{x ^{2} + 2}

e x t re m e f (x) = 2 π r^{2} + \frac{1500}{r}

e x t re m e f (x) = \frac{- 4}{x ^{2} - 1}

e x t re m e x^{3} + 2 x^{2} - x - 2

e x t re m e 5 - 7 x - 4 x^{2}