extreme 2x^3-3x^2-x+9

Lösung

extrempunkte

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

Maximum (\frac{3 - 15}{6}, \frac{5 15 - 18}{18} + 9), Minimum (\frac{3 + 15}{6}, \frac{- 18 - 5 15}{18} + 9)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} - 6 x - 1

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{5}{2 3} + \frac{1}{2},

Absteigend

: - \frac{5}{2 3} + \frac{1}{2} < x < \frac{5}{2 3} + \frac{1}{2},

Ansteigend

: \frac{5}{2 3} + \frac{1}{2} < x < \infty

Stecke

x = \frac{3 - 15}{6}

2 x^{3} - 3 x^{2} - x + 9 : \frac{5 15 - 18}{18} + 9

Maximum (\frac{3 - 15}{6}, \frac{5 15 - 18}{18} + 9)

Stecke

x = \frac{3 + 15}{6}

2 x^{3} - 3 x^{2} - x + 9 : \frac{- 18 - 5 15}{18} + 9

Minimum (\frac{3 + 15}{6}, \frac{- 18 - 5 15}{18} + 9)

Maximum (\frac{3 - 15}{6}, \frac{5 15 - 18}{18} + 9), Minimum (\frac{3 + 15}{6}, \frac{- 18 - 5 15}{18} + 9)

e x t re m e f (x) = (x - 6) e^{- 6 x}

e x t re m e 2 t^{10} (4 - t^{2})^{5}

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

e x t re m e f (y) = x^{3} + y^{3} - 27 x y

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