extreme 2x^3-x^2-4x+12

Lösung

extrempunkte

2 x^{3} - x^{2} - 4 x + 12

Maximum (- \frac{2}{3}, \frac{368}{27}), Minimum (1, 9)

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 + 12 : \frac{368}{27}

Maximum (- \frac{2}{3}, \frac{368}{27})

Stecke

x = 1

2 x^{3} - x^{2} - 4 x + 12 : 9

Minimum (1, 9)

Maximum (- \frac{2}{3}, \frac{368}{27}), Minimum (1, 9)

e x t re m e f (x) = x^{3} - \frac{3}{2} x^{2} - 6 x + 1

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

e x t re m e f (x) = 600 + 35 x

e x t re m e f (x) = 3 cos (π x), 0 \leq x \leq \frac{1}{6}

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