extreme f(x)=(t^2-4)^3

Lösung

extrempunkte

f (x) = (t^{2} - 4)^{3}

Sattel (- 2, 0), Minimum (0, - 64), Sattel (2, 0)

Schritte zur Lösung

f^{'} (t) = 6 t (t^{2} - 4)^{2}

Intervalle finden:

Absteigend

: - \infty < t < - 2,

Absteigend

: - 2 < t < 0,

Ansteigend

: 0 < t < 2,

Ansteigend

: 2 < t < \infty

Stecke

t = - 2

(t^{2} - 4)^{3} : 0

Sattel (- 2, 0)

Stecke

t = 0

(t^{2} - 4)^{3} : - 64

Minimum (0, - 64)

Stecke

t = 2

(t^{2} - 4)^{3} : 0

Sattel (2, 0)

Sattel (- 2, 0), Minimum (0, - 64), Sattel (2, 0)

e x t re m e \frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

e x t re m e f (x) = \frac{( x ^{2} - 5 x + 2 )}{x - 5}

e x t re m e (8 - x) (x + 1)^{2}

e x t re m e f (x) = x^{3} + 11 x + 10

e x t re m e f (x) = 3 sin (x), 0 \leq x \leq 2 π