extreme f(x)=(x^2-4)^7

Lösung

extrempunkte

f (x) = (x^{2} - 4)^{7}

Sattel (- 2, 0), Minimum (0, - 16384), Sattel (2, 0)

Schritte zur Lösung

f^{'} (x) = 14 x (x^{2} - 4)^{6}

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 0,

Ansteigend

: 0 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 2

(x^{2} - 4)^{7} : 0

Sattel (- 2, 0)

Stecke

x = 0

(x^{2} - 4)^{7} : - 16384

Minimum (0, - 16384)

Stecke

x = 2

(x^{2} - 4)^{7} : 0

Sattel (2, 0)

Sattel (- 2, 0), Minimum (0, - 16384), Sattel (2, 0)

e x t re m e f (x) = 2 x^{4} - 8 x + 3

e x t re m e 2 x^{3} + 3 x^{2} - 12 x + 5, 0 \leq x \leq 2

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 28 y + 261

e x t re m e f (x) = 24 x - 2 x^{3}

e x t re m e f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 5 x - 2 y