extreme f(x)=(x+6)(x-3)^2

Lösung

extrempunkte

f (x) = (x + 6) (x - 3)^{2}

Maximum (- 3, 108), Minimum (3, 0)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = - 3

(x + 6) (x - 3)^{2} : 108

Maximum (- 3, 108)

Stecke

x = 3

(x + 6) (x - 3)^{2} : 0

Minimum (3, 0)

Maximum (- 3, 108), Minimum (3, 0)

e x t re m e f (x) = 2 x^{3} - 24 x^{2} + 72 x + 10

e x t re m e f (x) = x^{2} - 2 x, 0 \leq x \leq 4

e x t re m e f (x) = x^{3} - x^{2} - 8 x + 8, - 2 \leq x \leq 0

roo t s 2 x y

e x t re m e f (x) = \frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9