extreme f(x)=x^3-2x^2-4x+12

Lösung

extrempunkte

f (x) = x^{3} - 2 x^{2} - 4 x + 12

Maximum (- \frac{2}{3}, \frac{364}{27}), Minimum (2, 4)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: 2 < x < \infty

Stecke

x = - \frac{2}{3}

x^{3} - 2 x^{2} - 4 x + 12 : \frac{364}{27}

Maximum (- \frac{2}{3}, \frac{364}{27})

Stecke

x = 2

x^{3} - 2 x^{2} - 4 x + 12 : 4

Minimum (2, 4)

Maximum (- \frac{2}{3}, \frac{364}{27}), Minimum (2, 4)

e x t re m e f (x) = - 5 cos (3 x) - 2

e x t re m e f (x) = x^{3} - 9 x^{2} + 24 x - 12

e x t re m e f (x) = \frac{x ^{3}}{3} - 6 x^{2} + 32 x

e x t re m e y^{2} - x^{2}

e x t re m e f (x) = 4 x - 5 x^{\frac{4}{5}}