extreme f(x)=2x^3-3x^2-6x

Lösung

extrempunkte

f (x) = 2 x^{3} - 3 x^{2} - 6 x

Maximum (\frac{1 - 5}{2}, - \frac{3 ( 3 - 5 )}{2} + 1 + 5), Minimum (\frac{1 + 5}{2}, - \frac{3 ( 3 + 5 )}{2} + 1 - 5)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

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

Stecke

x = \frac{1 - 5}{2}

2 x^{3} - 3 x^{2} - 6 x : - \frac{3 ( 3 - 5 )}{2} + 1 + 5

Maximum (\frac{1 - 5}{2}, - \frac{3 ( 3 - 5 )}{2} + 1 + 5)

Stecke

x = \frac{1 + 5}{2}

2 x^{3} - 3 x^{2} - 6 x : - \frac{3 ( 3 + 5 )}{2} + 1 - 5

Minimum (\frac{1 + 5}{2}, - \frac{3 ( 3 + 5 )}{2} + 1 - 5)

Maximum (\frac{1 - 5}{2}, - \frac{3 ( 3 - 5 )}{2} + 1 + 5), Minimum (\frac{1 + 5}{2}, - \frac{3 ( 3 + 5 )}{2} + 1 - 5)

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

e x t re m e f (x) = 0.00397 t^{2} - 0.593 t + 33.718

e x t re m e f (x) = f (x) = e x - x, - 4 \leq x \leq 2

e x t re m e f (x) = 4 x^{2} - 48 x + 190

e x t re m e f (x, y) = e^{9 x^{2} + 6 y^{2} + 1}