extreme f(x)=2x^3-3x^2-12x+12

Lösung

extrempunkte

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 12

Maximum (- 1, 19), Minimum (2, - 8)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} - 6 x - 12

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

: - 1 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 1

2 x^{3} - 3 x^{2} - 12 x + 12 : 19

Maximum (- 1, 19)

Stecke

x = 2

2 x^{3} - 3 x^{2} - 12 x + 12 : - 8

Minimum (2, - 8)

Maximum (- 1, 19), Minimum (2, - 8)

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

e x t re m e f (x, y) = e^{x} ln (y)

e x t re m e f (x) = 3 x^{4} - 4 x^{3}, - 1 \leq x \leq 2

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

e x t re m e f (x) = 2 x^{3} - 6 x^{2} - 48 x + 10