extreme f(x)=x^3(x+4)

Lösung

extrempunkte

f (x) = x^{3} (x + 4)

Minimum (- 3, - 27), Sattel (0, 0)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - 3

x^{3} (x + 4) : - 27

Minimum (- 3, - 27)

Stecke

x = 0

x^{3} (x + 4) : 0

Sattel (0, 0)

Minimum (- 3, - 27), Sattel (0, 0)

e x t re m e f (x, y) = x^{3} + y^{3} - 27 x y

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

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

e x t re m e f (x) = \frac{1}{9} \cdot (\frac{x ^{3}}{x + 2})

e x t re m e f (x) = x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4