extreme f(x)=2x^3-2x^2-2x+9

Lösung

extrempunkte

f (x) = 2 x^{3} - 2 x^{2} - 2 x + 9

Maximum (- \frac{1}{3}, \frac{253}{27}), Minimum (1, 7)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{1}{3}

2 x^{3} - 2 x^{2} - 2 x + 9 : \frac{253}{27}

Maximum (- \frac{1}{3}, \frac{253}{27})

Stecke

x = 1

2 x^{3} - 2 x^{2} - 2 x + 9 : 7

Minimum (1, 7)

Maximum (- \frac{1}{3}, \frac{253}{27}), Minimum (1, 7)

e x t re m e f (x) = x^{\frac{4}{7}} (3 x + 11)

e x t re m e 6 x + 18

e x t re m e f (x) = 5 x^{2} - x - 2

e x t re m e f (x) = x^{3} - 6 x^{2} + 5 [- 3.5]

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