extreme f(x)=60x^3-120x

Lösung

extrempunkte

f (x) = 60 x^{3} - 120 x

Maximum (- \frac{2}{3}, - 60 (\frac{2}{3})^{\frac{3}{2}} + 406), Minimum (\frac{2}{3}, 60 (\frac{2}{3})^{\frac{3}{2}} - 406)

Schritte zur Lösung

f^{'} (x) = 180 x^{2} - 120

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{2}{3},

Absteigend

: - \frac{2}{3} < x < \frac{2}{3},

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = - \frac{2}{3}

60 x^{3} - 120 x : - 60 (\frac{2}{3})^{\frac{3}{2}} + 406

Maximum (- \frac{2}{3}, - 60 (\frac{2}{3})^{\frac{3}{2}} + 406)

Stecke

x = \frac{2}{3}

60 x^{3} - 120 x : 60 (\frac{2}{3})^{\frac{3}{2}} - 406

Minimum (\frac{2}{3}, 60 (\frac{2}{3})^{\frac{3}{2}} - 406)

Maximum (- \frac{2}{3}, - 60 (\frac{2}{3})^{\frac{3}{2}} + 406), Minimum (\frac{2}{3}, 60 (\frac{2}{3})^{\frac{3}{2}} - 406)

e x t re m e 2 y

e x t re m e e^{x} (21 + 9 x - x^{2})

e x t re m e f (x) = x^{2} + y^{2} + x^{2} y + 7

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

e x t re m e f (x, y) = - 4 x^{2} - 8 y^{2} + 4 x - 16 y + 1