extreme f(x)=2x^4-2x^3

Lösung

extrempunkte

f (x) = 2 x^{4} - 2 x^{3}

Sattel (0, 0), Minimum (\frac{3}{4}, - \frac{27}{128})

Schritte zur Lösung

f^{'} (x) = 8 x^{3} - 6 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{3}{4},

Ansteigend

: \frac{3}{4} < x < \infty

Stecke

x = 0

2 x^{4} - 2 x^{3} : 0

Sattel (0, 0)

Stecke

x = \frac{3}{4}

2 x^{4} - 2 x^{3} : - \frac{27}{128}

Minimum (\frac{3}{4}, - \frac{27}{128})

Sattel (0, 0), Minimum (\frac{3}{4}, - \frac{27}{128})

e x t re m e \frac{x ^{2} + 1}{1 - x}

e x t re m e f (x) = 6 x^{3} - 9 x^{2} - 216 x + 3

e x t re m e f (x) = - x^{2} + 6, - 3 \leq x \leq 4

e x t re m e f (x) = 6 - 5 x - x^{3}

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