extreme f(x)=(x^3)/3-2x^2-5x-2

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} - 2 x^{2} - 5 x - 2

Maximum (- 1, \frac{2}{3}), Minimum (5, - \frac{106}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 4 x - 5

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

: - 1 < x < 5,

Ansteigend

: 5 < x < \infty

Stecke

x = - 1

\frac{x ^{3}}{3} - 2 x^{2} - 5 x - 2 : \frac{2}{3}

Maximum (- 1, \frac{2}{3})

Stecke

x = 5

\frac{x ^{3}}{3} - 2 x^{2} - 5 x - 2 : - \frac{106}{3}

Minimum (5, - \frac{106}{3})

Maximum (- 1, \frac{2}{3}), Minimum (5, - \frac{106}{3})

e x t re m e \frac{x}{x ^{2} - 64}

e x t re m e f (x) = 2 ln (x) + ln (y) - 4 x - y

e x t re m e f (x) = 20 e^{- 0.1 \cdot 10}

e x t re m e f (x) = ∣ x y ∣

e x t re m e f (x) = - 2 x^{2} + 12 - 25