extreme y=1+40x^3-3x^5

Lösung

extrempunkte

y = 1 + 40 x^{3} - 3 x^{5}

Minimum (- 22, 1 - 2562), Sattel (0, 1), Maximum (22, 1 + 2562)

Schritte zur Lösung

f^{'} (x) = - 15 x^{4} + 120 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < - 22,

Ansteigend

: - 22 < x < 0,

Ansteigend

: 0 < x < 22,

Absteigend

: 22 < x < \infty

Stecke

x = - 22

1 + 40 x^{3} - 3 x^{5} : 1 - 2562

Minimum (- 22, 1 - 2562)

Stecke

x = 0

1 + 40 x^{3} - 3 x^{5} : 1

Sattel (0, 1)

Stecke

x = 22

1 + 40 x^{3} - 3 x^{5} : 1 + 2562

Maximum (22, 1 + 2562)

Minimum (- 22, 1 - 2562), Sattel (0, 1), Maximum (22, 1 + 2562)

e x t re m e f (x) = (4, 2 x - 10) = (x - 1, y + 2)

e x t re m e f (x) = 80 x^{\frac{1}{4}} - x^{\frac{5}{4}}

e x t re m e f (x) = 120 t - 16 t^{2}

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

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