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

Lösung

extrempunkte

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

Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (1, - 1)

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 + 2 : \frac{98}{27}

Maximum (- \frac{2}{3}, \frac{98}{27})

Stecke

x = 1

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

Minimum (1, - 1)

Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (1, - 1)

e x t re m e f (x) = 2 x \cdot ln (x)

e x t re m e 6 x^{2} + 30 x - 84

e x t re m e y = r - (x - r)^{2}

e x t re m e f (x) = 2 x^{3} - 30 x^{2} + 96 x + 4, 0 \leq x \leq 12

e x t re m e f (x) = 3 x - x^{2} + 40