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

Lösung

extrempunkte

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

Maximum (- \frac{2}{3}, \frac{148}{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 + 4 : \frac{148}{27}

Maximum (- \frac{2}{3}, \frac{148}{27})

Stecke

x = 2

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

Minimum (2, - 4)

Maximum (- \frac{2}{3}, \frac{148}{27}), Minimum (2, - 4)

e x t re m e (x + 1)^{5} - 5 x - 2

e x t re m e f (x) = \frac{2 x - 1}{x ^{2}}

e x t re m e f (x, y) = (2 x - x^{2}) \cdot (2 y - y^{2})

e x t re m e 4 x - x^{2}

e x t re m e f (x) = - 5 x^{3} + 15 x + 8