extreme f(x)= 1/4 x^4+x^3-2

Lösung

extrempunkte

f (x) = \frac{1}{4} x^{4} + x^{3} - 2

Minimum (- 3, - \frac{35}{4}), Sattel (0, - 2)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - 3

\frac{1}{4} x^{4} + x^{3} - 2 : - \frac{35}{4}

Minimum (- 3, - \frac{35}{4})

Stecke

x = 0

\frac{1}{4} x^{4} + x^{3} - 2 : - 2

Sattel (0, - 2)

Minimum (- 3, - \frac{35}{4}), Sattel (0, - 2)

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

e x t re m e f (x) = 2 x^{3} - x^{2} - 20 x + 10

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

e x t re m e f (x, y) = x^{2} + 5 x y + y^{2} - 2 x + y - 6

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