extreme y=x^3-2x+4

Lösung

extrempunkte

y = x^{3} - 2 x + 4

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 4), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 4)

Schritte zur Lösung

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

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}

x^{3} - 2 x + 4 : - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 4

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 4)

Stecke

x = \frac{2}{3}

x^{3} - 2 x + 4 : (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 4

Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 4)

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 4), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 4)

e x t re m e f (x) = x^{2} + x y + y^{2} - 13 y + 56

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

e x t re m e f (x) = - x^{3} + 27 x - 62

e x t re m e P (a, c) = a + 7 + c

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