extreme f(x)=4x^3+48x^2+180x

Lösung

extrempunkte

f (x) = 4 x^{3} + 48 x^{2} + 180 x

Maximum (- 5, - 200), Minimum (- 3, - 216)

Schritte zur Lösung

f^{'} (x) = 12 x^{2} + 96 x + 180

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < - 3,

Ansteigend

: - 3 < x < \infty

Stecke

x = - 5

4 x^{3} + 48 x^{2} + 180 x : - 200

Maximum (- 5, - 200)

Stecke

x = - 3

4 x^{3} + 48 x^{2} + 180 x : - 216

Minimum (- 3, - 216)

Maximum (- 5, - 200), Minimum (- 3, - 216)

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

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

e x t re m e f (x, y) = x^{3} + y^{3} - 24 x y

e x t re m e f (x) = \frac{8}{x} + 2 π x^{2}

e x t re m e f (x, y) = 400 - 9 x^{2} - 36 y^{2}