extreme f(x)=x^3-6x^2+5[-3.5]

Lösung

extrempunkte

f (x) = x^{3} - 6 x^{2} + 5 [- 3.5]

Maximum (0, - 17.5), Minimum (4, - 49.5)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = 0

x^{3} - 6 x^{2} + 5 [- 3.5] : - 17.5

Maximum (0, - 17.5)

Stecke

x = 4

x^{3} - 6 x^{2} + 5 [- 3.5] : - 49.5

Minimum (4, - 49.5)

Maximum (0, - 17.5), Minimum (4, - 49.5)

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

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

e x t re m e f (x) = 2 x + \frac{20}{x}

e x t re m e f (x, y) = x^{3} + y^{2} - 6 x y - 29 x + 18 y + 20

e x t re m e f (x) = x^{2} - 6 x + 13