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

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{8}{5}, \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

Schritte zur Lösung

f^{'} (x) = - \frac{5 x - 8}{3 x ^{\frac{1}{3}}}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{8}{5},

Absteigend

: \frac{8}{5} < x < \infty

Stecke

x = 0

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

Minimum (0, 0)

Stecke

x = \frac{8}{5}

- x^{\frac{2}{3}} (x - 4) : \frac{48}{5 \cdot 5 ^{\frac{2}{3}}}

Maximum (\frac{8}{5}, \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

Minimum (0, 0), Maximum (\frac{8}{5}, \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

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

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

e x t re m e f (x) = x^{3} (x - 6)^{4}

e x t re m e f (x) = \frac{2}{3} x^{3} - 2 x^{2} + \frac{44}{3}

e x t re m e f (x) = 2 x^{3} - 24 x^{2} + 72 x + 7