extreme f(x)=-x^3+12x-8

Lösung

extrempunkte

f (x) = - x^{3} + 12 x - 8

Minimum (- 2, - 24), Maximum (2, 8)

Schritte zur Lösung

f^{'} (x) = - 3 x^{2} + 12

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = - 2

- x^{3} + 12 x - 8 : - 24

Minimum (- 2, - 24)

Stecke

x = 2

- x^{3} + 12 x - 8 : 8

Maximum (2, 8)

Minimum (- 2, - 24), Maximum (2, 8)

e x t re m e f (x) = x^{2} - 11, - 2 \leq x \leq 3

e x t re m e f (x y) = x^{3} + y^{2} - 6 x y + 9 x + 5 y + 2

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

e x t re m e f (x) = \frac{( 3 x - 6 )}{( x + 2 )}

e x t re m e f (x, y) = 3 x^{3} - y^{3} - 9 x + 12 y + 3