extreme f(x)=3x^2-4x+12

Lösung

extrempunkte

f (x) = 3 x^{2} - 4 x + 12

Minimum (\frac{2}{3}, \frac{32}{3})

Schritte zur Lösung

f^{'} (x) = 6 x - 4

Intervalle finden:

Absteigend

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

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = \frac{2}{3}

3 x^{2} - 4 x + 12 : \frac{32}{3}

Minimum (\frac{2}{3}, \frac{32}{3})

e x t re m e f (x) = x (- \frac{2 x}{3} + 4)

e x t re m e f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y - 6 x - 8 y + 11

e x t re m e f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 4 y

e x t re m e f (x) = 4 x^{5} - 25 x^{4} + 40 x^{3} - 4 [- 1.4]

e x t re m e f (x) = - \frac{1}{2} x^{2} + 15 x - 51