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

Lösung

extrempunkte

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

Maximum (- \frac{2}{3}, \frac{4}{27} - 12), Minimum (0, - 12)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

: - \frac{2}{3} < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - \frac{2}{3}

x^{3} + x^{2} - 12 : \frac{4}{27} - 12

Maximum (- \frac{2}{3}, \frac{4}{27} - 12)

Stecke

x = 0

x^{3} + x^{2} - 12 : - 12

Minimum (0, - 12)

Maximum (- \frac{2}{3}, \frac{4}{27} - 12), Minimum (0, - 12)

e x t re m e f (x) = x^{4} - 72 x^{2} + 1296, - 7 \leq x \leq 7

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

e x t re m e \frac{6 - x}{x ^{3}}

e x t re m e f (x, y) = 3 x y - 4 x - 4 y + 3

e x t re m e h (x) = 5 - x^{2} (- 3.1)