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

Lösung

extrempunkte

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

Maximum (\frac{8}{3}, \frac{32}{27}), Minimum (4, 0)

Schritte zur Lösung

f^{'} (x) = (x - 4) (3 x - 8)

Intervalle finden:

Ansteigend

: - \infty < x < \frac{8}{3},

Absteigend

: \frac{8}{3} < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = \frac{8}{3}

(x - 4)^{2} (x - 2) : \frac{32}{27}

Maximum (\frac{8}{3}, \frac{32}{27})

Stecke

x = 4

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

Minimum (4, 0)

Maximum (\frac{8}{3}, \frac{32}{27}), Minimum (4, 0)

e x t re m e x^{4} + 2 x^{3} + x^{2} + 2 x + 2

e x t re m e y = sin (x) + cos (x)

e x t re m e f (x) = 4 x^{2} + 40, - 1 \leq x \leq 1

e x t re m e f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

e x t re m e f (x) = 4 - x