extreme f(x)=-2/3 x^3+2x^2+60

Lösung

extrempunkte

f (x) = - \frac{2}{3} x^{3} + 2 x^{2} + 60

Minimum (0, 60), Maximum (2, \frac{188}{3})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = 0

- \frac{2}{3} x^{3} + 2 x^{2} + 60 : 60

Minimum (0, 60)

Stecke

x = 2

- \frac{2}{3} x^{3} + 2 x^{2} + 60 : \frac{188}{3}

Maximum (2, \frac{188}{3})

Minimum (0, 60), Maximum (2, \frac{188}{3})

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

e x t re m e y (t, β) = - 4 e^{- 4 t} (β + 15) + 5 e^{- 5 t} (β + 12)

e x t re m e f (x, y) = 2 x^{2} + 2 y^{2} - 4 y + 9

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

e x t re m e f (x) = x (x^{3} - x + 10)