extreme f(x)=x^3-2x^2-4x+1

Lösung

extrempunkte

f (x) = x^{3} - 2 x^{2} - 4 x + 1

Maximum (- \frac{2}{3}, \frac{67}{27}), Minimum (2, - 7)

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 + 1 : \frac{67}{27}

Maximum (- \frac{2}{3}, \frac{67}{27})

Stecke

x = 2

x^{3} - 2 x^{2} - 4 x + 1 : - 7

Minimum (2, - 7)

Maximum (- \frac{2}{3}, \frac{67}{27}), Minimum (2, - 7)

e x t re m e f (x, y) = 3 x^{2} + 2 y^{2} - 6 x - 4 y + 16

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

e x t re m e 2^{x}

e x t re m e y = (4 - x) 4^{x}

e x t re m e - 8 t^{2} + 36 t + 99