extreme f(x)=(x^2-1)^4

Lösung

extrempunkte

f (x) = (x^{2} - 1)^{4}

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

Schritte zur Lösung

f^{'} (x) = 8 x (x^{2} - 1)^{3}

Intervalle finden:

Absteigend

: - \infty < x < - 1,

Ansteigend

: - 1 < x < 0,

Absteigend

: 0 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - 1

(x^{2} - 1)^{4} : 0

Minimum (- 1, 0)

Stecke

x = 0

(x^{2} - 1)^{4} : 1

Maximum (0, 1)

Stecke

x = 1

(x^{2} - 1)^{4} : 0

Minimum (1, 0)

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

e x t re m e f (x) = x^{5} - 5 x^{4}

e x t re m e f (x) = x + \frac{1}{x}, 0.2 \leq x \leq 4

e x t re m e f (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} - 3 x^{2} + 11

e x t re m e f (x, y) = 7 x - 8 y + 2 x y - x^{2} + y^{3}

e x t re m e f (x) = - \frac{2}{3} x^{3} + 2 x^{2} + 6 x + 6