extreme f(x)=2x^3-x^2-4x+8(-1)

Lösung

extrempunkte

f (x) = 2 x^{3} - x^{2} - 4 x + 8 (- 1)

Maximum (- \frac{2}{3}, - \frac{172}{27}), Minimum (1, - 11)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} - 2 x - 4

Intervalle finden:

Ansteigend

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

Absteigend

: - \frac{2}{3} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{2}{3}

2 x^{3} - x^{2} - 4 x + 8 (- 1) : - \frac{172}{27}

Maximum (- \frac{2}{3}, - \frac{172}{27})

Stecke

x = 1

2 x^{3} - x^{2} - 4 x + 8 (- 1) : - 11

Minimum (1, - 11)

Maximum (- \frac{2}{3}, - \frac{172}{27}), Minimum (1, - 11)

e x t re m e f (x, y) = 2 x^{5} + x^{3} y^{2} + 6 x y^{4}

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

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

e x t re m e f (x) = 2 x^{4} - 20 x^{2} + 18

e x t re m e - x^{2} + 4 x + 6