extreme f(x)=3x^4-96x^2+1

Lösung

extrempunkte

f (x) = 3 x^{4} - 96 x^{2} + 1

Minimum (- 4, - 767), Maximum (0, 1), Minimum (4, - 767)

Schritte zur Lösung

f^{'} (x) = 12 x^{3} - 192 x

Intervalle finden:

Absteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < 0,

Absteigend

: 0 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = - 4

3 x^{4} - 96 x^{2} + 1 : - 767

Minimum (- 4, - 767)

Stecke

x = 0

3 x^{4} - 96 x^{2} + 1 : 1

Maximum (0, 1)

Stecke

x = 4

3 x^{4} - 96 x^{2} + 1 : - 767

Minimum (4, - 767)

Minimum (- 4, - 767), Maximum (0, 1), Minimum (4, - 767)

e x t re m e f (x) = 0.003 x^{2} + 4.2 x - 50

e x t re m e f (x) = - 3 x^{2} + 1

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

e x t re m e f (x) = 2 x^{3} + x^{2} - 20 x

e x t re m e f (x) = 2 x^{2} - 4 x [0.3]