extreme f(x)=x^4-4x^3+16x+4

Lösung

extrempunkte

f (x) = x^{4} - 4 x^{3} + 16 x + 4

Minimum (- 1, - 7), Sattel (2, 20)

Schritte zur Lösung

f^{'} (x) = 4 x^{3} - 12 x^{2} + 16

Intervalle finden:

Absteigend

: - \infty < x < - 1,

Ansteigend

: - 1 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 1

x^{4} - 4 x^{3} + 16 x + 4 : - 7

Minimum (- 1, - 7)

Stecke

x = 2

x^{4} - 4 x^{3} + 16 x + 4 : 20

Sattel (2, 20)

Minimum (- 1, - 7), Sattel (2, 20)

e x t re m e f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

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

e x t re m e P (x, y) = 6 x^{2} - x y - y^{2} + y + 8 x + 2

e x t re m e f (x) = (x + 2)^{\frac{4}{3}}

e x t re m e f (x) = x^{3} - 3 x^{2} + 1, - \frac{1}{2} \leq x \leq 4