extreme f(x)=x^4-12x^2+36

Lösung

extrempunkte

f (x) = x^{4} - 12 x^{2} + 36

Minimum (- 6, 0), Maximum (0, 36), Minimum (6, 0)

Schritte zur Lösung

f^{'} (x) = 4 x^{3} - 24 x

Intervalle finden:

Absteigend

: - \infty < x < - 6,

Ansteigend

: - 6 < x < 0,

Absteigend

: 0 < x < 6,

Ansteigend

: 6 < x < \infty

Stecke

x = - 6

x^{4} - 12 x^{2} + 36 : 0

Minimum (- 6, 0)

Stecke

x = 0

x^{4} - 12 x^{2} + 36 : 36

Maximum (0, 36)

Stecke

x = 6

x^{4} - 12 x^{2} + 36 : 0

Minimum (6, 0)

Minimum (- 6, 0), Maximum (0, 36), Minimum (6, 0)

e x t re m e f (x) = 1 + cos (x)

e x t re m e f (x) = x^{2} e^{- 6 x}

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

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

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