extreme f(x)=3x^2+4x-6

Lösung

extrempunkte

f (x) = 3 x^{2} + 4 x - 6

Minimum (- \frac{2}{3}, - \frac{22}{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 - 6 : - \frac{22}{3}

Minimum (- \frac{2}{3}, - \frac{22}{3})

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

e x t re m e f (x, y) = 7 e^{y} - 3 y e^{x}

e x t re m e \frac{1}{t ^{3} + 2}

e x t re m e f (x) = \frac{6}{7} (x^{2} - 9)^{\frac{2}{3}}

e x t re m e f (x, y) = 7 x y + 14 x - x^{2} + 2 y^{2}