extreme f(x)=(x^3)/3-x^2-8x-4

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} - x^{2} - 8 x - 4

Maximum (- 2, \frac{16}{3}), Minimum (4, - \frac{92}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 2 x - 8

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = - 2

\frac{x ^{3}}{3} - x^{2} - 8 x - 4 : \frac{16}{3}

Maximum (- 2, \frac{16}{3})

Stecke

x = 4

\frac{x ^{3}}{3} - x^{2} - 8 x - 4 : - \frac{92}{3}

Minimum (4, - \frac{92}{3})

Maximum (- 2, \frac{16}{3}), Minimum (4, - \frac{92}{3})

e x t re m e f (x) = x^{4} - 5 x^{3} + 1.1 x^{2} - 10.2 x + 1

e x t re m e (x - 1) (x + 9)^{2}

e x t re m e x^{\frac{1}{5}} (x^{2} - 4)

e x t re m e f (x) = cos (\frac{4 x}{3})

e x t re m e \frac{5 x}{1 - x}