extreme f(x)=3x^3+3x^2-10x-2

Lösung

extrempunkte

f (x) = 3 x^{3} + 3 x^{2} - 10 x - 2

Maximum (- \frac{1 + 11}{3}, \frac{32 + 22 11}{9} - 2), Minimum (\frac{- 1 + 11}{3}, \frac{32 - 22 11}{9} - 2)

Schritte zur Lösung

f^{'} (x) = 9 x^{2} + 6 x - 10

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 11 - 1}{3},

Absteigend

: \frac{- 11 - 1}{3} < x < \frac{11 - 1}{3},

Ansteigend

: \frac{11 - 1}{3} < x < \infty

Stecke

x = - \frac{1 + 11}{3}

3 x^{3} + 3 x^{2} - 10 x - 2 : \frac{32 + 22 11}{9} - 2

Maximum (- \frac{1 + 11}{3}, \frac{32 + 22 11}{9} - 2)

Stecke

x = \frac{- 1 + 11}{3}

3 x^{3} + 3 x^{2} - 10 x - 2 : \frac{32 - 22 11}{9} - 2

Minimum (\frac{- 1 + 11}{3}, \frac{32 - 22 11}{9} - 2)

Maximum (- \frac{1 + 11}{3}, \frac{32 + 22 11}{9} - 2), Minimum (\frac{- 1 + 11}{3}, \frac{32 - 22 11}{9} - 2)

e x t re m e f (x) = 0.0001 x^{2} + 3.2 x - 90

e x t re m e f (x, y) = θ

e x t re m e \frac{x ^{3}}{3} - x^{2} - 35 x + 4

e x t re m e f (x, y) = e^{6 x^{2} + 4 y^{2} + 10}

e x t re m e f (x, y) = x y - 8 x - y^{2} + 12 x + 160