extreme f(x)=(6x+1)e^{8x-x^2}

Lösung

extrempunkte

f (x) = (6 x + 1) e^{8 x - x^{2}}

Minimum (\frac{23 - 697}{12}, e^{\frac{491 - 25 697}{72}} (\frac{23 - 697}{2} + 1)), Maximum (\frac{23 + 697}{12}, e^{\frac{491 + 25 697}{72}} (\frac{23 + 697}{2} + 1))

Schritte zur Lösung

f^{'} (x) = - 2 e^{- x (x - 8)} (6 x^{2} - 23 x - 7)

Intervalle finden:

Absteigend

: - \infty < x < \frac{23 - 697}{12},

Ansteigend

: \frac{23 - 697}{12} < x < \frac{23 + 697}{12},

Absteigend

: \frac{23 + 697}{12} < x < \infty

Stecke

x = \frac{23 - 697}{12}

(6 x + 1) e^{8 x - x^{2}} : e^{\frac{491 - 25 697}{72}} (\frac{23 - 697}{2} + 1)

Minimum (\frac{23 - 697}{12}, e^{\frac{491 - 25 697}{72}} (\frac{23 - 697}{2} + 1))

Stecke

x = \frac{23 + 697}{12}

(6 x + 1) e^{8 x - x^{2}} : e^{\frac{491 + 25 697}{72}} (\frac{23 + 697}{2} + 1)

Maximum (\frac{23 + 697}{12}, e^{\frac{491 + 25 697}{72}} (\frac{23 + 697}{2} + 1))

Minimum (\frac{23 - 697}{12}, e^{\frac{491 - 25 697}{72}} (\frac{23 - 697}{2} + 1)), Maximum (\frac{23 + 697}{12}, e^{\frac{491 + 25 697}{72}} (\frac{23 + 697}{2} + 1))

e x t re m e f (x) = x^{4} - 3 x^{2}, - 2 \leq x \leq 0

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

e x t re m e y = 15 x e^{- 0.05 x} - 30

e x t re m e f (x) = \frac{- 4 x}{x ^{2} + 6}, (0, 6)

e x t re m e f (x) = 2 x^{3} + 15 x^{2} - 36 x + 7, - 6 < x < 2