extreme f(x)=x^4(x-3)^3

Lösung

extrempunkte

f (x) = x^{4} (x - 3)^{3}

Maximum (0, 0), Minimum (\frac{12}{7}, - \frac{15116544}{823543}), Sattel (3, 0)

Schritte zur Lösung

f^{'} (x) = x^{3} (x - 3)^{2} (7 x - 12)

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{12}{7},

Ansteigend

: \frac{12}{7} < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = 0

x^{4} (x - 3)^{3} : 0

Maximum (0, 0)

Stecke

x = \frac{12}{7}

x^{4} (x - 3)^{3} : - \frac{15116544}{823543}

Minimum (\frac{12}{7}, - \frac{15116544}{823543})

Stecke

x = 3

x^{4} (x - 3)^{3} : 0

Sattel (3, 0)

Maximum (0, 0), Minimum (\frac{12}{7}, - \frac{15116544}{823543}), Sattel (3, 0)

e x t re m e f (x) = (x - 1)^{2} + (4 - 4 x^{2})

e x t re m e f (x) = x^{3} + y^{3} - 30 x y

e x t re m e f (x) = \frac{5 x}{x + 1}

e x t re m e y = \frac{x ^{3} ( 3 x - 8 )}{12}

e x t re m e f (x) = e^{3 x} (7 - x)