extreme f(x)=8x^2-2x^4

Lösung

extrempunkte

f (x) = 8 x^{2} - 2 x^{4}

Maximum (- 2, 8), Minimum (0, 0), Maximum (2, 8)

Schritte zur Lösung

f^{'} (x) = 16 x - 8 x^{3}

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 0,

Ansteigend

: 0 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = - 2

8 x^{2} - 2 x^{4} : 8

Maximum (- 2, 8)

Stecke

x = 0

8 x^{2} - 2 x^{4} : 0

Minimum (0, 0)

Stecke

x = 2

8 x^{2} - 2 x^{4} : 8

Maximum (2, 8)

Maximum (- 2, 8), Minimum (0, 0), Maximum (2, 8)

e x t re m e P (q, s) = q + 2 + s

e x t re m e f (x) = 4 x^{2} e^{- 0.25 x}

e x t re m e f (x) = x^{3} - 9 x^{2} + 15 x + 2

e x t re m e f (x) = - \frac{2}{3} x^{3} + 9 x^{2} - 28 x + 1

e x t re m e f (x) = ln (x^{2} + 4)