extreme f(x)=(x^2-12)*e^{-2x}

Lösung

extrempunkte

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

Minimum (- 3, - 3 e^{6}), Maximum (4, \frac{4}{e ^{8}})

Schritte zur Lösung

f^{'} (x) = - 2 e^{- 2 x} (x + 3) (x - 4)

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 4,

Absteigend

: 4 < x < \infty

Stecke

x = - 3

(x^{2} - 12) \cdot e^{- 2 x} : - 3 e^{6}

Minimum (- 3, - 3 e^{6})

Stecke

x = 4

(x^{2} - 12) \cdot e^{- 2 x} : \frac{4}{e ^{8}}

Maximum (4, \frac{4}{e ^{8}})

Minimum (- 3, - 3 e^{6}), Maximum (4, \frac{4}{e ^{8}})

e x t re m e f (x) = 49 x^{2} - 10 x + 17

e x t re m e y = x^{7} e^{x} + 4

e x t re m e 3 x^{3} + 30 x^{2} + 75 x - 9

e x t re m e f (x, y) = x^{3} + y^{3} - 63 (x + y) + 12 x y

e x t re m e g (x) = \frac{1}{9} x^{9} - x