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

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{12}{5}, \frac{839808}{3125}), Sattel (6, 0)

Schritte zur Lösung

f^{'} (x) = - x (x - 6)^{2} (5 x - 12)

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{12}{5},

Absteigend

: \frac{12}{5} < x < 6,

Absteigend

: 6 < x < \infty

Stecke

x = 0

x^{2} (6 - x)^{3} : 0

Minimum (0, 0)

Stecke

x = \frac{12}{5}

x^{2} (6 - x)^{3} : \frac{839808}{3125}

Maximum (\frac{12}{5}, \frac{839808}{3125})

Stecke

x = 6

x^{2} (6 - x)^{3} : 0

Sattel (6, 0)

Minimum (0, 0), Maximum (\frac{12}{5}, \frac{839808}{3125}), Sattel (6, 0)

e x t re m e f (x) = x^{2} + x y + y^{2} - 31 y + 320

e x t re m e f (x, y) = - 2 x^{3} - 2 y^{3} - 6 x y + 1

e x t re m e P (x, y) = 3 x^{2} - 8 x y - 3 y^{2}

e x t re m e f (x) = x^{4} - 2 x^{2} + 8

e x t re m e f (x) = - \frac{1}{x - 2}