extreme f(x)=15x^4-4x^3

Lösung

extrempunkte

f (x) = 15 x^{4} - 4 x^{3}

Sattel (0, 0), Minimum (\frac{1}{5}, - \frac{1}{125})

Schritte zur Lösung

f^{'} (x) = 60 x^{3} - 12 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{1}{5},

Ansteigend

: \frac{1}{5} < x < \infty

Stecke

x = 0

15 x^{4} - 4 x^{3} : 0

Sattel (0, 0)

Stecke

x = \frac{1}{5}

15 x^{4} - 4 x^{3} : - \frac{1}{125}

Minimum (\frac{1}{5}, - \frac{1}{125})

Sattel (0, 0), Minimum (\frac{1}{5}, - \frac{1}{125})

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

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

e x t re m e f (x, y) = x^{3} - 12 x - y^{2} - 8 y + 18

e x t re m e f (x) = 2 x^{3} - 9 x^{2} + 12 x + 7

e x t re m e f (x) = 2 x^{3} - 9 x^{2} + 12 x + 1