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

Lösung

extrempunkte

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

Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{3456}{3125}), Sattel (2, 0)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{4}{5},

Ansteigend

: \frac{4}{5} < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = 0

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

Maximum (0, 0)

Stecke

x = \frac{4}{5}

x^{2} (x - 2)^{3} : - \frac{3456}{3125}

Minimum (\frac{4}{5}, - \frac{3456}{3125})

Stecke

x = 2

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

Sattel (2, 0)

Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{3456}{3125}), Sattel (2, 0)

e x t re m e f (x) = \frac{6 x ^{5}}{5} - 15 x^{3} - 81 x + 2

e x t re m e - 0.1 t^{2} 0.8 t + 98.8

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

e x t re m e f (x) = cos (π x), 0 \leq x \leq \frac{1}{6}

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