extreme f(x)=x^3-12x^3

Lösung

extrempunkte

f (x) = x^{3} - 12 x^{3}

Sattel (0, 0)

Schritte zur Lösung

f^{'} (x) = - 33 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \infty

Stecke

x = 0

x^{3} - 12 x^{3} : 0

Sattel (0, 0)

e x t re m e x^{3} + 2 x^{2} + x + 1

e x t re m e f (x y) = 3 x^{2} - 5 x y + 2 y^{2} + 14 x + 7 y - 5

e x t re m e f (x) = \frac{2 ( x + 7 ) ^{2}}{x + 10}, (- 10, - 6)

e x t re m e f (x, y) = x^{6} - 3 x^{2} y^{2} + 3 x y^{3}

e x t re m e - x^{3} + x^{2} + 8 x + 8