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

Lösung

extrempunkte

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

Maximum (- \frac{1}{2}, 2 + 6), Minimum (\frac{1}{2}, - 2 + 6)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1}{2},

Absteigend

: - \frac{1}{2} < x < \frac{1}{2},

Ansteigend

: \frac{1}{2} < x < \infty

Stecke

x = - \frac{1}{2}

2 x^{3} - 3 x + 6 : 2 + 6

Maximum (- \frac{1}{2}, 2 + 6)

Stecke

x = \frac{1}{2}

2 x^{3} - 3 x + 6 : - 2 + 6

Minimum (\frac{1}{2}, - 2 + 6)

Maximum (- \frac{1}{2}, 2 + 6), Minimum (\frac{1}{2}, - 2 + 6)

e x t re m e \frac{2 x - 3}{x}

e x t re m e f (x) = x + 3 (1 - x)^{\frac{1}{3}}

e x t re m e 10 + x^{2} + 6 x + 10

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

e x t re m e 3 + 6 x^{2} - 4 x^{3}