extreme f(x)=-x^3-6x^2-1

Lösung

extrempunkte

f (x) = - x^{3} - 6 x^{2} - 1

Minimum (- 4, - 33), Maximum (0, - 1)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < 0,

Absteigend

: 0 < x < \infty

Stecke

x = - 4

- x^{3} - 6 x^{2} - 1 : - 33

Minimum (- 4, - 33)

Stecke

x = 0

- x^{3} - 6 x^{2} - 1 : - 1

Maximum (0, - 1)

Minimum (- 4, - 33), Maximum (0, - 1)

e x t re m e f (x) = \frac{x ^{2} - 4}{x ^{2} + 4}, - 5 \leq x \leq 5

e x t re m e y = - 4 x^{2} - 5 x + 1

e x t re m e f (x) = \frac{x ^{2} - 7 x - 6}{x - 5}

e x t re m e (x + 1)^{3} (x - 3)

e x t re m e f (x) = 2 x^{3} + 9 x^{2} - 60 x + 6, - 5 \leq x \leq 5