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

Lösung

extrempunkte

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

Minimum (0, 2), Maximum (2, 6)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = 0

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

Minimum (0, 2)

Stecke

x = 2

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

Maximum (2, 6)

Minimum (0, 2), Maximum (2, 6)

e x t re m e f (x) = \frac{1}{x} + ln (x), 0.5 \leq x \leq 4

e x t re m e f (x) = x^{2} + 6 x + 6

e x t re m e f (x, y) = 2 x^{3} - 3 x^{2} y - 12 x^{2} - 3 y^{2}

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

e x t re m e \frac{x ^{3}}{x + 1}