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

Lösung

extrempunkte

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}})

Schritte zur Lösung

f^{'} (x) = e^{7 x} \cdot 7 - e^{- x}

Intervalle finden:

Absteigend

: - \infty < x < - \frac{1}{8} ln (7),

Ansteigend

: - \frac{1}{8} ln (7) < x < \infty

Stecke

x = - \frac{ln ( 7 )}{8}

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}})

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

e x t re m e f (x) = x 81 - x^{2}

e x t re m e S (a, d) = a + (n - 1) d

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

e x t re m e y = 3 x + 2 cos (x), 0 \leq x \leq 2 π