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

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{27})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{4}{3},

Absteigend

: \frac{4}{3} < x < \infty

Stecke

x = 0

2 x^{2} - x^{3} : 0

Minimum (0, 0)

Stecke

x = \frac{4}{3}

2 x^{2} - x^{3} : \frac{32}{27}

Maximum (\frac{4}{3}, \frac{32}{27})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{27})

e x t re m e f (x) = - x^{2} + 6 x + 1, 5 \leq x \leq 8

e x t re m e f (x) = x^{3} + 2 x^{2} - 4 x + 8

e x t re m e f (x, y) = x^{2} + y^{2} - 81

e x t re m e f (x, y) = 11 x^{2} + 4 x y + 5 y^{2} + 5 x + 5 y

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