extreme x^{11}-3x^9+2

Lösung

extrempunkte

x^{11} - 3 x^{9} + 2

Maximum (- \frac{3 3}{11}, \frac{9565938 3}{161051 11} + 2), Sattel (0, 2), Minimum (\frac{3 3}{11}, - \frac{9565938 3}{161051 11} + 2)

Schritte zur Lösung

f^{'} (x) = 11 x^{10} - 27 x^{8}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{3 3}{11},

Absteigend

: - \frac{3 3}{11} < x < 0,

Absteigend

: 0 < x < \frac{3 3}{11},

Ansteigend

: \frac{3 3}{11} < x < \infty

Stecke

x = - \frac{3 3}{11}

x^{11} - 3 x^{9} + 2 : \frac{9565938 3}{161051 11} + 2

Maximum (- \frac{3 3}{11}, \frac{9565938 3}{161051 11} + 2)

Stecke

x = 0

x^{11} - 3 x^{9} + 2 : 2

Sattel (0, 2)

Stecke

x = \frac{3 3}{11}

x^{11} - 3 x^{9} + 2 : - \frac{9565938 3}{161051 11} + 2

Minimum (\frac{3 3}{11}, - \frac{9565938 3}{161051 11} + 2)

Maximum (- \frac{3 3}{11}, \frac{9565938 3}{161051 11} + 2), Sattel (0, 2), Minimum (\frac{3 3}{11}, - \frac{9565938 3}{161051 11} + 2)

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

e x t re m e 9 x^{3} - 7 x^{2} + 3 x + 10

e x t re m e f (x) = 4 + 27 x - x^{3}

e x t re m e f (x) = - \frac{8 x}{( x ^{2} + 1 ) ^{2}}

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