extreme f(x)=-1/3 x^3+2x^2-3x-12

Lösung

extrempunkte

f (x) = - \frac{1}{3} x^{3} + 2 x^{2} - 3 x - 12

Minimum (1, - \frac{40}{3}), Maximum (3, - 12)

Schritte zur Lösung

f^{'} (x) = - x^{2} + 4 x - 3

Intervalle finden:

Absteigend

: - \infty < x < 1,

Ansteigend

: 1 < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = 1

- \frac{1}{3} x^{3} + 2 x^{2} - 3 x - 12 : - \frac{40}{3}

Minimum (1, - \frac{40}{3})

Stecke

x = 3

- \frac{1}{3} x^{3} + 2 x^{2} - 3 x - 12 : - 12

Maximum (3, - 12)

Minimum (1, - \frac{40}{3}), Maximum (3, - 12)

e x t re m e f (x, y) = 2 x^{4} + y^{2} - 16 x y

e x t re m e f (x, y) = 2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 4

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

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

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