extreme f(x)=x^3-12x^2+36x-1

Lösung

extrempunkte

f (x) = x^{3} - 12 x^{2} + 36 x - 1

Maximum (2, 31), Minimum (6, - 1)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 24 x + 36

Intervalle finden:

Ansteigend

: - \infty < x < 2,

Absteigend

: 2 < x < 6,

Ansteigend

: 6 < x < \infty

Stecke

x = 2

x^{3} - 12 x^{2} + 36 x - 1 : 31

Maximum (2, 31)

Stecke

x = 6

x^{3} - 12 x^{2} + 36 x - 1 : - 1

Minimum (6, - 1)

Maximum (2, 31), Minimum (6, - 1)

e x t re m e f (x) = x - \frac{16}{3} x^{3}

e x t re m e f (x) = x^{4} - 8 x^{3} + 16 x^{2} - 10

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

e x t re m e f (x) = - 3 x^{4} + 6 x^{2} + 1

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