extreme (x+4)^3(3x-2)^2

Lösung

extrempunkte

(x + 4)^{3} (3 x - 2)^{2}

Sattel (- 4, 0), Maximum (- \frac{6}{5}, \frac{2151296}{3125}), Minimum (\frac{2}{3}, 0)

Schritte zur Lösung

f^{'} (x) = 3 (x + 4)^{2} (3 x - 2) (5 x + 6)

Intervalle finden:

Ansteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < - \frac{6}{5},

Absteigend

: - \frac{6}{5} < x < \frac{2}{3},

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = - 4

(x + 4)^{3} (3 x - 2)^{2} : 0

Sattel (- 4, 0)

Stecke

x = - \frac{6}{5}

(x + 4)^{3} (3 x - 2)^{2} : \frac{2151296}{3125}

Maximum (- \frac{6}{5}, \frac{2151296}{3125})

Stecke

x = \frac{2}{3}

(x + 4)^{3} (3 x - 2)^{2} : 0

Minimum (\frac{2}{3}, 0)

Sattel (- 4, 0), Maximum (- \frac{6}{5}, \frac{2151296}{3125}), Minimum (\frac{2}{3}, 0)

e x t re m e f (x) = 3 x^{2} - 12

e x t re m e f (x, y) = x^{2} + x y + \frac{1}{2} y^{2} - 2 x + y

e x t re m e f (x) = 14 x^{2} - 2 x^{3} 2 y^{2} + 4 x y

e x t re m e 2 x^{2} + y^{2} + x y - 3 x

e x t re m e f (x, y) = 4 x^{2} - x y + 4 y^{2}