extreme f(x)=24x-24x^2+6x^3

Lösung

extrempunkte

f (x) = 24 x - 24 x^{2} + 6 x^{3}

Maximum (\frac{2}{3}, \frac{64}{9}), Minimum (2, 0)

Schritte zur Lösung

f^{'} (x) = 24 - 48 x + 18 x^{2}

Intervalle finden:

Ansteigend

: - \infty < x < \frac{2}{3},

Absteigend

: \frac{2}{3} < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = \frac{2}{3}

24 x - 24 x^{2} + 6 x^{3} : \frac{64}{9}

Maximum (\frac{2}{3}, \frac{64}{9})

Stecke

x = 2

24 x - 24 x^{2} + 6 x^{3} : 0

Minimum (2, 0)

Maximum (\frac{2}{3}, \frac{64}{9}), Minimum (2, 0)

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

e x t re m e f (x) = x^{2} lo g_{9} (x)

e x t re m e f (x) = 2000 + 530 (ln (x))

e x t re m e f (x) = \frac{3 x + 4}{6 x + 1}

e x t re m e f (x) = 3 x^{4} - 96 x^{2} + 1