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

Lösung

extrempunkte

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

Maximum (- \frac{2}{3}, \frac{4 2}{9} + 1), Minimum (\frac{2}{3}, - \frac{4 2}{9} + 1)

Schritte zur Lösung

f^{'} (x) = 9 x^{2} - 2

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{2}{3}

3 x^{3} - 2 x + 1 : \frac{4 2}{9} + 1

Maximum (- \frac{2}{3}, \frac{4 2}{9} + 1)

Stecke

x = \frac{2}{3}

3 x^{3} - 2 x + 1 : - \frac{4 2}{9} + 1

Minimum (\frac{2}{3}, - \frac{4 2}{9} + 1)

Maximum (- \frac{2}{3}, \frac{4 2}{9} + 1), Minimum (\frac{2}{3}, - \frac{4 2}{9} + 1)

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

e x t re m e f (x) = x^{4} - 8 x^{3} + 10 x^{2} + 1

e x t re m e - 15 x^{4} + 120 x^{2}

e x t re m e 3 (x - 2)^{2} (x - 1) + (x - 2)^{2}

e x t re m e f (x, y) = \frac{3 x - y}{4}