extreme (x+2)^2(x-3)

Lösung

extrempunkte

(x + 2)^{2} (x - 3)

Maximum (- 2, 0), Minimum (\frac{4}{3}, - \frac{500}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < \frac{4}{3},

Ansteigend

: \frac{4}{3} < x < \infty

Stecke

x = - 2

(x + 2)^{2} (x - 3) : 0

Maximum (- 2, 0)

Stecke

x = \frac{4}{3}

(x + 2)^{2} (x - 3) : - \frac{500}{27}

Minimum (\frac{4}{3}, - \frac{500}{27})

Maximum (- 2, 0), Minimum (\frac{4}{3}, - \frac{500}{27})

e x t re m e f (x) = 3 cos (2 x) + 4

e x t re m e f (x) = 3 x + 6 x^{- 1}

e x t re m e f (x, y) = - x^{3} + y^{3} + 15 x y + 1

e x t re m e f (x, y) = 5 x^{2} y - 3 x y^{2} - x^{2} - y

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